Answer:
Earth.
Explanation:
Center of gravity can be defined as the specific point where all of the weight of an object is concentrated.
Generally, all the objects found around the world all have a center of gravity.
When an object is balanced so that a displacement lowers its center of gravity, the object is said to be in stable equilibrium.
Hence, the force of gravity acting on an object is directed through this center of gravity and toward the center of the earth.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, weight is given by the formula;
[tex] Weight, W = mg [/tex]
Where;
m is the mass of an object.
g is acceleration due to gravity.
In the laboratory, a ball is dropped onto a force-sensing platform several times, each time hitting a different surface (foam, feathers, clay, etc.). The momentum of the ball changes by the same amount in each trial; in each trial, the average scale reading is F, and the time of collision t are measured. What quantities would need to be graphed to exhibit a straight-line relationship
Answer:
Graphing the momentum against the change in moment yields a linear relationship.
Explanation:
This is an impulse experiment,
I = ∫ F .dt
where the force and time of the collision are measured, therefore if we assume an average force the integral reduces to
I = F t
Furthermore, the momentum is equal to the change in moment of the ball, this change in moment can be found using the energy relations measuring the height of the ball and calculating its speed, in the two intervals for the descent and for the exit, possibly the heights are different so the moment change is different from zero.
Starting point. Higher
Em₀ = U = mgh
Lower end point, just before hitting the scale
[tex]Em_{f}[/tex] = K = ½ m v²
in the path in the air there is no friction
Em₀ = Em_{f}
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
this height is different for the descent and ascent of the ball, so we have two moments
Δp = [tex]p_{f}[/tex] - p₀
Δp = m (v_{f} -v₀)
therefore we have the relationship
I = Δp
Graphing the momentum against the change in moment yields a linear relationship.
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
Which type of heat transfer causes air movement between land and ocean?
There are three methods of heat travel:
CONDUCTION -- The transfer of heat through a medium. This is how we cook food on top of a stove. The heat from the stove burner is conducted through a medium (a metal pot) to the food.
CONVECTION -- The transfer of heat due to the physical movement of an object. We can observe convection by looking at a pot of boiling water. Have you ever noticed that when a pot of water is boiling, the water seems to follow a vertical circular motion? This is convection. The parcel of heated water at the bottom of the pot rises, as it rises it gives off some of its heat. Because it loses some heat, the parcel is cooler than the surrounding water. It then sinks to the bottom of the pot and the process is started again. The path of the rising water followed by the sinking water traces out a circle.
RADIATION -- The transfer of heat by means of waves. This is the most difficult method of heat transfer to understand. Yet, we experience it every day. We feel the effects of radiation whenever we stand near a stove or oven which is being used. We feel the heat radiating from the stove or oven to our skin. Similarly, we have all been outside on a sunny, hot Summer's day. If we look up to the sky we can feel the rays of the Sun hitting our faces. The Sun is radiating its heat to the Earth.
It is through one of the above processes of heat transfer that causes the air temperature at deep-ocean station 41001 to be warmer than that of land station CLKN7 during the winter months. Which process do you believe to be the cause of the air temperature differences between these two stations? I'll give you a hint, it has something to do with the temperature of the ocean water. Lets look at a graph of both the average air and water temperatures from Station 41001.
As you can see from the graph, the January (month 1) and February (month 2) water temperatures are about 20 degrees while the respective air temperatures are about 15 degrees. This is a 5 degree difference in temperature between the air and the water at the same geographical location!!
We can figure out what heat transfer process is influencing the air temperature at station 41001 by applying the three methods to our situation and then we can choose the one that seems most logical.
First, lets look at conduction. This process involves the transfer of heat through a conductive medium. Well, nothing exists between the air and the water surface. In our situation, the heat is going directly from the water to the air without passing through a conductive medium. Therefore, this is not the applicable process that is causing the warm winter-time air temperatures at station 41001.
Convection involves the movement of heated objects. The physical movement must be a result of the heating, such as with the pot of boiling water where the vertical movement is caused by the intense heat applied to the bottom of the pot. Because the ocean water isn't moving into or through the atmosphere as a result of the sun's heating of the water, convection isn't the process influencing air and water temperature difference. Ocean water is moving through the lower few feet of the air as ocean surface waves, but this doesn't occur because of the sun's heat.
The final process, radiation, is causing the winter-time air temperatures over water to be warmer than the winter-time air temperatures over land. The heat of the ocean is being given off (radiated) into the air, thus making the air substantially warmer.
An Egyptian pyramid contains approximately 1.95 million stone blocks. The average weight of each block is 2.55 tons. What is the weight of the pyramid in pounds?
Answer:
More than 2,300,000 limestone and granite blocks were pushed, pulled, and dragged into place on the Great Pyramid. The average weight of a block is about 2.3 metric tons (2.5 tons).
28. Which of the following correctly shows the order of highest amount of friction to the lowest amount of
friction?
a. Static, Rolling, Sliding
b. Sliding, Rolling, Static
c. Rolling, Static, Sliding
d. Static, Sliding, Rolling
Answer:
[tex]\mathrm{d.\:Static,\: Sliding,\:Rolling}[/tex]
Explanation:
Static friction occurs when an object initially starts at rest. When the surfaces of the materials touch, the microscopic unevenness interlock greatest with each other, causing the most friction out of the three.
During sliding friction, an object is already moving or in motion. The microscopic surfaces still interlock, but because the object is in motion, it has a momentum. Therefore, the magnitude of sliding friction is less than that of static friction.
Rolling friction occurs when an object rolls across some surface. Rather than surfaces interlocking, rolling friction is caused by the constant distortion of surfaces. As it rolls, the surfaces of the object are constantly wrapping and changing. This distortion causes the rolling friction. However, it is much less in magnitude when compared to static or sliding friction.
Question 1 of 10
What might happen to personal information when it is transferred using
digital signals?
A. Some information might be changed when the data are copied.
B. It might be accessed by someone who was not the intended
recipient.
C. The information might change while being transmitted because of
noise.
D. The information might change to analog, making it less reliable.
Answer:
its b for sure
Explanation:
Answer:
B. It might be accessed by someone who was not the intended
recipient
a student lifts a 15N mass through a distance of 1.5m. whats the works done ?
Answer:
10N/m
Explanation:
Calculating workdone=Force/Distance
Therefore=15N/1.5m
=10N/m
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
*PLEASE HELP*
When an object is placed in front
of a convex lens, it creates a virtual
image at -12.8 cm with a
magnification of 2.85. What is the
focal length of the lens?
(Mind your minus signs.)
(Unit = cm)
Answer: The focal length of the lens = 3.32 cm.
Explanation:
Lens formula : [tex]\dfrac1f=\dfrac1v-\dfrac1u[/tex] (i)
f= focal length , v=image distance , u =object distance.
magnification: m = [tex]\dfrac{v}{u}[/tex] (ii )
Given: v= -12.8 cm , m =2.85
Put values in (ii), we get
[tex]2.85=\dfrac{-12.8}{u}\\\\\Rightarrow\ u=\dfrac{-12.8}{2.85}\\\\\Rightarrow\ u=-4.49\ cm[/tex]
substitute values of u , v in (i)
[tex]\dfrac1f=\dfrac1{-12.8}-\dfrac{1}{4.49}\\\\\Rightarrow\dfrac1f=-0.30084214922\\\\\Rightarrow\ f=\dfrac{-1}{-0.30084214922}\approx3.32\ cm[/tex]
Hence, the focal length of the lens = 3.32 cm.
Answer:
6.92
Explanation:
2.85=-(-12.8/x)
do=4.49
1/f= 1/4.49 + 1/-12.8
f=6.92
I need help with number 3!!!!!!!!!!!!
Answer:
3. The frequency of the wave is 3 Hz.
Explanation:
3. Determination of the frequency of the wave.
Frequency is simply defined as the number of complete circle or oscillation made in 1 seconds. Mathematically, it can be expressed as:
f = n / t
Where:
f => is the frequency.
n => is the number of circle.
t => is the time.
With the above formula, we can obtain the frequency of the wave as follow:
Number of complete circle (n) = 3
Time (t) = 1 s
Frequency (f) =?
f = n / t
f = 3 / 1
f = 3 /s = 3 Hz
Therefore, the frequency of the wave is 3 Hz
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
Firemen are shooting a stream of water at a burning building. A high-pressure hose shoots out the water with a speed of 26.0 m/s as it leaves the hose nozzle. Once it leaves the hose, the water moves in projectile motion. The firemen adjust the angle of elevation of the hose until the water takes 3.00 to reach a building 41.0m away. You can ignore air resistance; assume that the end of the hose is at ground level.
Required:
a. Find the angle of elevation of the hose.
b. Find the speed in m/s of the water at the highest point in its trajectory.
c. Find the acceleration in m/s^2 of the water at the highest point in its trajectory.
d. How high above the ground in m does the water strike the building?
e. How fast is it moving in m/s just before it hits the building?
Answer:
a) θ = 58.3º
b) vfh = 13.7 m/s
c) g = -9.8 m/s2
d) h = 22.2 m
e) vfb = 15.5 m/s
Explanation:
a)
Assuming that gravity is the only influence that causes an acceleration to the water, due to it is always downward, since both directions are independent each other, in the horizontal direction, the water moves at a constant speed.Since the velocity vector has a magnitude of 26.0 m/s, we can find its horizontal component as follows:vₓ₀ = v * cos θ (1)where θ is the angle between the water and the horizontal axis (which we define as the x-axis, being positive to the right).Applying the definition of average velocity, taking the end of the hose like the origin, and making t₀ = 0, we can write the following expression:[tex]x_{f} = v_{ox} * t = v_{o} * cos \theta * t (2)[/tex]
Replacing by the givens of xf = 41.0m, t = 3.00 s, and v=26.0 m/s, we can solve for the angle of elevation θ, as follows:[tex]cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)[/tex]
⇒θ = cos⁻¹ (0.526) = 58.3º (4)b)
At the highest point in its trajectory, just before starting to fall, the vertical component of the velocity is just zero.Since the horizontal component keeps constant during all the journey, we can conclude that the speed at this point is just v₀ₓ, that we can find easily from (1) replacing by the values of v and cos θ, as follows:vₓ₀ = v * cos θ = 26.0 m/s * 0.526 = 13.7 m/s. (5)c)
At any point in the trajectory, the only acceleration present is due to the action of gravity, which accepted value is -9.8 m/s2 (taking the upward direction on the vertical y-axis as positive)d)
Since we know the time when the water strikes the building, it will be the same for the vertical movement, so, we can use the kinematic equation for vertical displacement, as follows:[tex]\Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)[/tex]
Our only unknown remains v₀y, which can be obtained in the same way than the horizontal component:v₀y = v * sin θ = 26.0 m/s * 0.85 = 22.1 m/s (7)Replacing (7) in (6), we get:[tex]\Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)[/tex]
e)
When the water hits the building the velocity vector, has two components, the horizontal vₓ and the vertical vy.The horizontal component, since it keeps constant, is just v₀x:v₀ₓ = 13.7 m/sThe vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:[tex]v_{fy} = v_{oy} - g*t (9)[/tex]
Replacing by the time t (a given), g, and v₀y from (7), we can solve (9) as follows:[tex]v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s (10)[/tex]
Since we know the values of both components (perpendicular each other), we can find the magnitude of the velocity vector (the speed, i.e. how fast is it moving), applying the Pythagorean Theorem to v₀ₓ and v₀y, as follows:[tex]v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)[/tex]
A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it strike the ground?
Answer:
[tex]t = 1.82[/tex]
Explanation:
Given
[tex]u = 7.70m/s[/tex] -- initial velocity
[tex]s = 30.2m[/tex] --- height
Required
Determine the time to hit the ground
This will be solved using the following motion equation.
[tex]s = ut + \frac{1}{2}gt^2[/tex]
Where
[tex]g = 9,8m/s^2[/tex]
So, we have:
[tex]30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2[/tex]
[tex]30.2 = 7.70t + 4.9 * t^2[/tex]
Subtract 30.2 from both sides
[tex]30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2[/tex]
[tex]0 = 7.70t + 4.9 * t^2 - 30.2[/tex]
[tex]0 = 7.70t + 4.9t^2 - 30.2[/tex]
[tex]7.70t + 4.9t^2 - 30.2 = 0[/tex]
[tex]4.9t^2 + 7.70t - 30.2 = 0[/tex]
Solve using quadratic formula:
[tex]t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}[/tex]
Where
[tex]a = 4.9;\ b = 7.70;\ c = -30.2[/tex]
[tex]t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}[/tex]
[tex]t = \frac{-7.70\±\sqrt{651.21}}{9.8}[/tex]
[tex]t = \frac{-7.70\±25.52}{9.8}[/tex]
Split the expression
[tex]t = \frac{-7.70+25.52}{9.8}[/tex] or [tex]t = \frac{-7.70-25.52}{9.8}[/tex]
[tex]t = \frac{17.82}{9.8}[/tex] or [tex]t = -\frac{33.22}{9.8}[/tex]
Time can't be negative; So, we have:
[tex]t = \frac{17.82}{9.8}[/tex]
[tex]t = 1.82[/tex]
Hence, the time to hit the ground is 1.82 seconds
Two spherical objects with a mass of 3.17 kg each are placed at a distance of 2.96 m apart. How many electrons need to leave each object so that the net force between them becomes zero
Answer:
1.704 × 10¹⁹ electrons
Explanation:
The gravitational force due to the identical masses with mass, m = 3.17 kg at a distance r = 2.96 m from each other is F = Gm²/r².
Since equal number of electrons have to leave both masses, we have a charge q on each mass acting to oppose each other at distance r.
So, the electrical force of repulsion is thus
F'= kq²/r²
For the net force to be zero, the gravitational force of attraction must balance the electrical force of repulsion.
So, F = F'
So, Gm²/r² = kq²/r²
Gm² = kq²
q² = Gm²/k
taking square root of both sides, we have
q = m√(G/k)
So, substituting the values of the variables into the equation, we have
q = m√(G/k)
= 3.17 kg√(6.67 × 10⁻¹¹ Nm²/kg²/9 × 10⁹ Nm²/C²)
= 3.17 kg√(0.741 × 10⁻²⁰ C²/kg²)
= 3.17 kg × 0.861 C/kg
= 2.73 C
Now, q = ne where n = number of electrons and e = electron charge = 1.602 × 10⁻¹⁹ C
n = q/e
= 2.73 C ÷ 1.602 × 10⁻¹⁹ C
= 1.704 × 10¹⁹ electrons
What are the three concepts of The Cell Theory?
Answer:
These findings led to the formation of the modern cell theory, which has three main additions: first, that DNA is passed between cells during cell division; second, that the cells of all organisms within a similar species are mostly the same, both structurally and chemically; and finally, that energy flow occurs within.
Explanation:
hope this helped! <3
can i have the crown ?
A basketball of mass 0.23kg is thrown horizontally against a rigid vertical wall with a velocity of 20m/s. It rebounds with a velocity of 15m/s. Calculate the impulse of the force of the wall on the basketball.
Answer:
[tex]8.1\:\mathrm{Ns}[/tex]
Explanation:
The impulse-momentum theorem gives the impulse on an object to be equal to the change in momentum of that object. Since mass is maintained, the change in momentum of the basketball is:
[tex]\Delta p = m\Delta v[/tex], where [tex]m[/tex] is the mass of the basketball and [tex]\Delta v[/tex] is the change in velocity.
Since the basketball is changing direction, its total change in velocity is:
[tex]\Delta v = 20-(-15)=35\:\mathrm{m/s}[/tex].
Therefore, the basketball's change in momentum is:
[tex]\Delta p = m\Delta v = 0.23\cdot 35= 8.05=8.1\:\mathrm{kg\cdot m/s}[/tex].
Thus, the impulse on the basketball is [tex]\fbox{$8.1\:\mathrm{Ns}$}[/tex] (two significant figures).
An air-filled parallel-plate capacitor is charged and then disconnected from the battery. The plates are then pulled apart to twice their original separation. Which of the following statements about this capacitor is true?
A. The capacitance has doubled.
B. The energy stored in the capacitor has doubled.
C. The potential difference across the plates has decreased.
D. The electric field between the plates has increased.
Answer:
B.
Explanation:
From the given information, since the capacitors are disconnected from the battery, the electric field between the plate does not change due to the fact that there is no difference in charge density.
Hence, the potential difference after separating the plates is:
V' = Ed
V = e (2d')
V' = 2V
For the energy stored in the capacitor;
[tex]U' = \dfrac{1}{2}QV'[/tex]
where;
V' = 2V
[tex]U' = \dfrac{1}{2} Q (2V)}[/tex]
[tex]U' =2( \dfrac{1}{2} Q V)[/tex]
U' = 2U
Thus, in the capacitor, the energy that is being stored is doubled.
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, how much money, to the nearest dollar, would be in the account after 6 years?
Answer:
15448
Explanation:
A=11000(1.01425)^{24}
A=11000(1.01425)
24
Austin invested $11,000 in an account paying an interest rate of 5.7% compounded quarterly. Assuming no deposits or withdrawals are made, the money to the nearest dollar, would be in the account after 6 years is 15448.
What is Compound interest?The compound interest occurs when the interest is reinvested rather than paying it out. It's basically earning interest over interest.
The formula is:
Compound interest, [tex]A = P ( 1 +\frac{r}{n} )^{nt}[/tex]
Where:
A = final Amount
P = initial principal balance
r = interest rate
n = number of times interest applied per time period
t = number of time periods elapsed
Austin invested P=$11000 in an account with an interest rate of r=5.7% = 0.057 (decimal) during t=6 years compounded quarterly. Since there are 4 quarters in a year, n=4.
Thus, Substituting all the values in the given formula,
A = 11000 ( 1 + [tex]\frac{0.057}{4} )^{6*4}[/tex]
= 11000 × 1.4043662796
= 15448.0290
The money to the nearest dollar, would be in the account after 6 years is 15448.
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John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
Find the value of F1 + F2 + F3.
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
For F₁
[tex]F_{y}=2[N][/tex]
For F₂
[tex]F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N][/tex]
For F₃
[tex]F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N][/tex]
Now we can sum each one of the forces in the given axes:
[tex]F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N][/tex]
Now using the Pythagorean theorem we can find the total force.
[tex]F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N][/tex]
1. An engine absorbs 600 J of heat while doing 650 J of work. What is the change
in internal energy of the enginge? *
1250 J
-50 J
-1250 J
50 J
The change in internal energy of the engine is -50 joule. Hence, option (B) is correct.
What is law of conservation of energy?Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.
The absorb energy: Q = 600 Joule
Work done: W = 650 Joule.
Let, the change in internal energy of the engine= dU.
According to conservation of energy:
The absorb energy = change in internal energy + Work done
Q = dU + W
dU = Q - W
= 600 joule - 650 joule
= - 50 joule.
Hence, the change in internal energy of the engine is -50 joule.
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a toy car has a 2.0 A current, and its resistance is 1.75 ohms. How much voltage does the car require
Answer:
the answer will be 24.40 ohms law
Explanation:
Which has a greater momentum and greater kinetic energy-a truck with
a mass of 3530 kg moving at a speed of 21 m/s or a car with a mass of
1620 kg moving at a speed of 54 m/s? Answers below are given in the
form: greater momentum, greater kinetic energy.
Answer:
the car has greater momentum.
the car has greater kinetic energy.
Explanation:
FOR MOMENTUM:
Momentum is given as the product of mass and velocity of an object:
[tex]P = mv[/tex]
where,
P = momentum
m = mass
v = velocity
For Truck:
[tex]P_{truck} = (3530\ kg)(21\ m/s)\\P_{truck} = 74130 Ns[/tex]
For Car:
[tex]P_{car} = (1620\ kg)(54\ m/s)\\P_{car} = 87480\ Ns[/tex]
Therefore, car has greater momentum.
FOR KINETIC ENERGY:
Kinetic Energy is given as:
[tex]K.E = \frac{1}{2} mv^{2}[/tex]
where,
K.E = Kinetic Energy
m = mass
v = velocity
For Truck:
[tex]K.E_{truck} = \frac{1}{2} (3530\ kg)(21\ m/s)^{2}\\K.E_{truck} = 778365\ J = 778.36\ KJ[/tex]
For Car:
[tex]K.E_{truck} = \frac{1}{2} (1620\ kg)(54\ m/s)^{2}\\K.E_{truck} = 2361960\ J = 2361.96\ KJ[/tex]
Therefore, car has greater kinetic energy.
A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is the corresponding raft speed?
Answer:
The answer is below
Explanation:
Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.
Momentum = mass * velocity
The principle of conservation of momentum states that momentum cannot be created or destroyed but can be transferred. Therefore the momentum before and after an action is equal.
Initial momentum = Final momentum
Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.
m = 71 kg, M = 500 kg, v = 6 m/s
Initial both the raft and diver are at rest, hence u and U is zero, hence:
mu + MU = mv + MV
71(0) + 500(0) = 71(6) + 500(V)
0 = 426 + 500(V)
500(V) = -426
V = -426/500
V = -0.852 m/s
. Dan drags a box across the floor. He uses 95 N of force and moves the box 12 m. How much work does he do? 0.13 J 1,140 J 7.9 J 107 J
Answer:
[tex]1140\:\mathrm{J}[/tex]
Explanation:
Work is given by [tex]W=F\Delta x[/tex], where [tex]F[/tex] is force and [tex]\Delta x[/tex] is displacement.
Plugging in given values, we get:
[tex]W=95\:\mathrm{N}\cdot12\:\mathrm{m}=\fbox{$1140\:\mathrm{J}$}[/tex].
Can someone please help meee .
Answer:
32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice
5.
What is the apparent colour of a red shirt when viewed in pure green light.?
Red
(b)- Green
Yellow (d) Black) (e) Blue
Answer: black
Explanation: When green light is shone on a red object, it absorbs all of the green light and not reflecting anything. Hence, it appears black.
A motor is used to lift a 10 kg mass 2 m above the ground in 4 s. If the power input to the motor is 100 W. what is the efficiency of the motor?
(Show Work)
Answer:
98%
Explanation:
Given parameters
Mass of motor = 10kg
Height = 2m
Time = 2s
Power input = 100w
Unknown
Efficiency = ?
Solution
Efficiency is the percentage of the power output to the power input.
Power is the rate at which work is done.
Power output = mass x g x height / time
g is the acceleration due to gravity
Power output = 10x 2 x 9.8 / 2 = 98W
Efficiency = power output/ power input x 100
Efficiency = 98/100 x 100 = 98%
A charge of 7.1 x 10-4 C is placed at the origin of a Cartesian coordinate system. A second charge of 6.5 x 10-4 C lies 20 cm above the origin, and a third charge of 8.9 x 10-4 C lies 20 cm to the right of the origin. Determine the direction of the total force on the first charge at the origin. Express your answer as a positive angle in degrees measured counter clockwise from the positive x-axis.
Answer:
α = 36.21 °
β = 143.79°
Explanation:
To do this, we need to know the expression to calculate the angle.
In this case:
α₁ = tan⁻¹ (Fy₁/Fx₁) (1)
Now, let's analize the given data.
We have a charge q₁ at the origin of the cartesian coordinate system, so, it's at the 0. The charge q₂ is 20 cm above q₁, meaning is on the y-axis. Finally q₃ it's 20 cm to the right, meaning it's on the x-axis.
Knowing this,we can calculate the force that q₂ and q₃ are exerting over q₁. As these forces are in the x and y-axis respectively, we also are calculating the value of the forces in the x and y axis, that are needed to calculate the direction.
The expression to calculate the force would be Coulomb's law so:
F = K q₁q₂ / r² (2)
The value of K is 9x10⁹ N m² / C². Let's calculate the forces:
F₁₂ = Fy = 9x10⁹ * (7.1x10⁻⁴) * (6.5x10⁻⁴) / (0.020)²
Fy = 1.04x10⁷ N
F₁₃ = Fx = 9x10⁹ * (7.1x10⁻⁴) * (8.9x10⁻⁴) / (0.020)²
Fx = 1.42x10⁷ N
Now that we have both forces, we can calculate the magnitude of the force:
F = √(Fx)² + (Fy)²
F = √(1.04x10⁷)² + (1.42x10⁷)²
F = 1.76x10⁷ N
Finally, the direction would be applying (1):
α = tan⁻¹ (1.04x10⁷/1.42x10⁷)
α = 36.21 °
And counter clockwise it would be:
β = 180 - 36.21 = 143.79°
Hope this helps