Answer:
A
Explanation:
Finding Gravitational Potential Energy :
GPE = mass x gravitational acceleration x height
GPE = 50 x 10⁻² x 10 x 2.5GPE = 5 x 2.5GPE = 12.5 JFinding Kinetic Energy :
KE = 1/2 x mass x (velocity)²
KE = 1/2 x 5 x 10⁻² x (0)²KE = 0Hence, option A is correct.
When you throw a ball into the air, its kinetic energy _
Answer:
increases
Explanation:
The ball's kinetic energy is the energy it has due to its motion. When you throw the ball into the air, it gains kinetic energy from the force of your throw.
in previous years. you were taught about the three basic food groups. in this activity, classify the focos into GO, GROW or GLOW
There are three basic food groups. The foods are classified into Go, Grow and Glow.
What is a GO food?Go Foods are the type of food that provide fuel and help us 'go' and be active whole day. 'Go' foods are: bread, rice, pasta, cereals and potato.
Grow foods are the ones which makes us grow by giving muscle strength and purifies blood.
Glow foods are the one which make us glow by face and skin healthy.
Foods classified into the three groups:
Toast+ Eggs + Spinach : GO -GROW-GLOW
oatmeal + nuts + berries : GO -GROW-GLOW
stir fried tofu + broccoli + rice : GROW-GLOW -GO
toast + apples + peanut butter: GO -GROW-GLOW
crackers+ hummus + cucumber :GO -GROW-GLOW
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X-ray production occurs in which process?
Answer:
Explanation:
X-rays are produced due to sudden deceleration of fast-moving electrons when they collide and interact with the target anode. In this process of deceleration, more than 99% of the electron energy is converted into heat and less than 1% of energy is converted into x-rays
(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of the forces to assist you with the calculation.)
(8)Figure 5 shows three charges arranged in a right angled formation.
(8.1)Draw a free body diagram of the forces that act on the -0,03 uC charge.
(8.2)Calculate each force that acts on the -0,03 uC charge.
(8.3) Find the magnitude and direction of the net force that acts on the 0,03 μC charge with the aid of a diagram and by calculations.
Help Please.
Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is
[tex]F = \dfrac{kq_1q_2}{r^2}[/tex]
where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :
[tex]F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}[/tex]
Force due to Q₃ :
[tex]F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}[/tex]
8.3. The net force on the particle at Q₂ is the vector
[tex]\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}[/tex]
Its magnitude is
[tex]\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}[/tex]
and makes an angle θ with the positive horizontal axis (pointing to the right) such that
[tex]\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}[/tex]
where we subtract 180° because [tex]\vec F[/tex] terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
You are on a boat in the middle of the Pacific Ocean at the equator traveling in a hydrofoil
going at a constant speed of 300 /. The water is perfectly still. What is your acceleration:
a) If you’re heading due North?
b) If you’re heading due East?
c) If you’re heading straight up (something probably went wrong at this point).
You may assume the following:
The earth has a radius of 6371 km.
The earth makes one full revolution every 24 hours.
The gravitational constant at sea level is 9.81 m/s2
East and North are relative to the Earth’s axial north, not magnetic north.
a) If you’re heading due North, total acceleration will be 0.0478 m/s².
b) If you’re heading due East, acceleration will be 0.0803 m/s².
c) If you’re heading straight up, no exact value for acceleration.
What is acceleration?The time rate of change of velocity is known as acceleration.
a)If the boat is going along North from equator, there will be only centrifugal force acting along upward direction which will be balanced by the weight of boat.
Angular velocity ω= (2π / (3600 x 24)) = 7.27 x 10⁻⁵ rad/s
Acceleration due to centrifugal force a = 0.0337m/s²
Centrifugal force due to its own velocity 300 m/s is v²/r = 0.0141 m/s².
Total acceleration will be 0.0141 +0.0337 =0.0478m/s².
Thus, if you’re heading due North, total acceleration will be 0.0478 m/s².
b) If the boat goes along East, Coriolis acceleration is in the direction of centrifugal force.
Coriolis acceleration is (2 x 7.27 x 10⁻⁵ x300 ) = 0.0436m/s².
Total centrifugal acceleration due to centrifugal force = 300 + (7.27 x 10⁻⁵ x 6371 x 10³) = 0.0436 m/s².
Total acceleration is 0.0436 + 0.0367 = 0.0803m/s².
Thus, if you’re heading due East, acceleration will be 0.0803 m/s².
c) When boat is going upward, centrifugal force will change with time. but Coriolis acceleration remains same.
Thus, if you’re heading straight up, no exact value for acceleration.
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A 1.0KW kettle contains 500g of boiling water. Calculate the time needed to evaporate all the water in the kettle. (Specific latent heat of vaporization of water=2.26MJKG^-1
The time needed to evaporate all the water in the kettle is 1.13 x 10³s.
What is Power?The power is defined as the the ratio of amount of energy produced per unit time.
P =E/t
Energy is in the form of heat gained to evaporate all water in kettle.
E = Q =mL
So, P = mL/t
Substitute mass of water m = 0.5kg, latent heat of vaporization of wate L = 2.26 x 10⁶ J/kg, power P= 10³ W, we have
10³ = (0.5 x 2.26 x 10⁶) /t
t = 1.13 x 10³s
Thus, the time needed to evaporate all the water is 1.13 x 10³s.
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The iron rim is made slightly smaller than the wooden wheel. How is this rim fitted on the wooden wheel?
BEST ANSWER WILL BE MARKED AS BRAINLIEST
Answer:
The rim is usually heated before fixing into the wooden wheel because on heating the iron rim, it expands first, then on cooling contracts and fits onto the wooden wheel.
Explanation: Hope it helps you :))))
Have a good day
What is the unit of electrical power?
Answer:
Watts
Explanation:
A Watt is the unit of electrical power equal to one ampere under the pressure of one volt.
Rajani had bought a new bottle of pickle from the market. She tried to open the
metal cap but could not do so. She then took a bowl of hot water and immersed
the upper end of the bottle in it for five minutes. She could easily open the
bottle now. Why?
BEST ANSWER WILL BE MARKED
Answer:
The pickle bottle cap on dipping it in hot water expanded.
Explanation: Hope it helps you:))))
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A certain car traveling at 97 km/h can stop in 47 m on a level road find the coefficient of friction
The coefficient of friction between the road and the car's tire is determined as 0.78.
Acceleration of the carThe acceleration of the car is calculated as follows;
v² = u² - 2as
0 = u² - 2as
a = u²/2s
where;
u is the initial velocity = 97 km/h = 26.94 m/sa = (26.94)²/(2 x 47)
a = 7.72 m/s²
Coefficient of frictionμ = a/g
μ = (7.72)/9.8
μ = 0.78
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A 425-g piece of metal at 100°C is dropped into a 100-g aluminum cup containing 500 g of water at 15°C. The final temperature of the system is 40°C. What is the specific heat of the metal, assuming no heat is exchanged with the surroundings? The specific heat of aluminum is 900 J/kg°C.
a. 2140 J/(kg•K)
b. 3800 J/(kg•K)
c. 3300 J/(kg•K)
d. 1900 J/(kg•K)
e. 4280 J/(kg•K)
The specific heat of the metal, assuming no heat is exchanged with the surroundings is 2140 J/(kg•K).
Specific heat capacity of the metal
The specific heat capacity of the metal is determined from the principle of conservation of energy.
energy lost by the metal = energy gained by aluminum + energy gained by water
Q = mcΔθ
where;
m is mass (kg)c is specific heat capacityΔθ is change in temperature0.425c(100 - 40) = 0.1(900)(40 - 15) + 0.5(4186)(40 - 15)
25.5c = 2250 + 52,325
c = 54,575/25.5
c = 2140 J/(kg•K)
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The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^24 kg. What is the gravitational force between these two objects?
Answer:
3 820 885 N
Explanation:
Gravitational equation
F = G m1 m2 / r^2
G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2
F = 6.6713 x 10^-11 * 4.41 x 10^5 * 5.97 x 10^24 / ( 6.78x 10^6)^2
= 3820885 .3 N
wood is a better insulator than glass so why is fiberglass commonly used as an insulator in wooden buildings why?
Answer:
So that it may trap the solar radiations and maintain the temperature of the wooden building.
Explanation:
Wood is not as effective an insulator as fiberglass. It comes in a number of widths, the widest being 3/4 inch, which has an R-value of 0.94. However, fiberglass, the thinnest of which is 3/4 inch, has an R-value of 3. Therefore, fiberglass is a better insulator.
An object is placed 4.0 cm to the left of a convex lens with a focal length of +8.0 cm . Where is the image of the object?
The image of the object is 8cm to the left of the lens (D)
What is the image of an object?
The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.
It is calculated thus:
1÷v = 1÷f - 1÷u
How to calculate the image of an objectFrom the formula
1÷v = 1÷f - 1÷u
WhereV = image distance fromthe object
U = object
f = focal length
Substitute the values
1÷v = 1÷8 - 1÷ 4
1÷v = - 1÷8
Make v the subject of formula
v = -8cm
Therefore, the image of the object is 8cm to the left of the lens (D)
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Two identical masses are attached to frictionless pulleys by very light strings wrapped around the rim of the pulley and are released from rest. Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop As the masses fall in which case is the tension in the string greateror is it the same in both cases? Justify your answer
The tension in the string is same in both cases because of the similar mass or weight of the bodies.
Is tension be the same in both cases?Both pulleys have the same mass and same diameter, but one is solid and the other is a hoop, then there is same tension occurs in both cases because tension depends on mass which is similar on both side. We know that tension is the opposite force of the weight.
So we can conclude that the tension in the string is same in both cases because of the similar mass or weight of the bodies.
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with detailed explaniation
#A
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=20sin^237[/tex]
[tex]\\ \rm\Rrightarrow H_{max}=7.2m[/tex]
#B
[tex]\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}[/tex]
[tex]\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}[/tex]
[tex]\\ \rm\Rrightarrow R=40sin74[/tex]
[tex]\\ \rm\Rrightarrow R=38.5m[/tex]
#C
[tex]\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}[/tex]
[tex]\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}[/tex]
[tex]\\ \rm\Rrightarrow T=4sin37[/tex]
[tex]\\ \rm\Rrightarrow T=2.4s[/tex]
Now
[tex]\\ \rm\Rrightarrow v=u-gt[/tex]
[tex]\\ \rm\Rrightarrow v=20-10(2.4)[/tex]
[tex]\\ \rm\Rrightarrow v=20-24[/tex]
[tex]\\ \rm\Rrightarrow v=-4m/s[/tex]
A bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s. What is the angular acceleration of the tires?
The angular acceleration of a bike with tires of radius 0.330 m speeds up from rest to 5.33 m/s in 6.27 s will be 2. 57 rad/s²
What is angular acceleration?
Angular acceleration can be defined as the time it takes for a change in angular velocity. It is denoted as 'α' with a unit of rads/s²
It is expressed thus;
α= Δω ÷ Δt
Where α = angular acceleration
Δw = change in angular velocity = velocity ÷ radius
Δ t = change in time
How to calculate the angular accelerationUsing the formula:
α= Δω ÷ Δt
v = 5. 33m/s, r = 0. 33m and t = 6.27s
Substitute the values to get the angular velocity
Δw = v÷ r = 5. 33 ÷ 0.330 = 16. 15 m/s
Substitute the value of Δw into the equation
α= Δω ÷ Δt = 16. 15 ÷ 6. 27 = 2. 57 rad/s²
Therefore, the angular acceleration of a bike with tires radius of 0. 330m, speed of 5. 33mls in 6. 27s is 2. 57 rad/s²
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What is the equivalent resistance of a circuit that contains two 50.00
resistors connected in series with a 12.0 V battery?
Ο A. 0.120 Ω
Ο Β. 100.0 Ω
OC. 0.480
OD. 25.00
Answer:
B
Explanation:
Resistors in series add because there is only one path the current can take.
calculations
= R1 + R2
= 50 + 50
= 100
discribe how standard of meter,second, and kilogram are agreed internationally.
Answer:
Hindi ko Po alm sorry Po talaga
How does the period of a wave change if the frequency increases by a factor of three?
Answer:
Period becomes 1/3 of the original
Explanation:
Period = 1/f if f becomes 3f then period becomes 1/3 of the original
What is the importance of science to your life?
Answer:
The beauty of science is everywhere.
Explanation:
science is extremely important and you see it all around you. For instance you need knowledge of chemistry to create medication in order to help cure the sick. Knowledge of physics is needed in order to construct building or vehicles.
make sure to ask if you need anymore examples or proof :)
The rubber band contains .......potential energy as it is stretched.
less
more
same
Answer:
elastic potential energy
You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.
Explanation:
edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement in the westerly direction?
The displacement of Edward in the westerly direction is determined as 180.27 km.
Displacement of EdwardThe displacement of Edward is calculated as follows;
R² = a² + b² - 2abcosθ
R² = 150² + 200² - 2(150 x 200) x cos60
R² = 32500
R = 180.27 km
Thus, the displacement of Edward in the westerly direction is determined as 180.27 km.
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a pool ball is rolling along a table3 with a constant velocity. the components of its velocity vector are Vx=0.5 m/s and Vy=0.8 m/s. Calculate the distance it travels in 0.4 s
The distance traveled by the pool ball is 0.376 m.
To calculate the distance traveled by the ball, we need to first find the resultant velocity of the ball.
What is resultant velocity?This is the single velocity obtained when two or more velocities are combined.
To calculate the resultant velocity, we use the formula below.
Formula:
Vr = √(Vx²+Vy²)............. Equation 1Where:
Vr = Resultant velocity.From the question,
Given:
Vx = 0.5 m/sVy = 0.8 m/sSubstitute these values into equation 1
Vr = √(0.5²+0.8²)Vr = √(0.25+0.64)Vr = √(0.89)Vr = 0.94 m/sFinally, to calculate the total distance, we use the formula below.
Formula:
d = (Vr)t.......... Equation 2Where:
d = Distance traveled by the ballt = timeFrom the question,
Given:
Vr = 0.94 m/st = 0.4 sSubstitute into equation 2
d = 0.4×0.94d = 0.376 mHence, the distance traveled by the pool ball is 0.376 m.
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at what temperature does the fahrenheit scale read quartered the celsius scale ? what is the corresponding reading in kelvin scale ?
The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C
Reading in kelvin scale = 252.5 K
Determine the temperature at which the Fahrenheit scale reads quartered the celsius scaleThe relationship between Fahrenheit and celsius can be expressed as
F = (9/5)(C) + 32 ---- ( 1 )
Therefore the temperature at which Fahrenheit scale quarters the celsius scale is :
F = 1/4 C --- ( 2 )
Back to equation ( 1 )
1/4 C = (9/5)(C) + 32
1/4C - 9/5C = 32
C ( 1/4 - 9/5 ) = 32
C = 32 / ( 1/4 - 9/5 )
= -20.64⁰C
F = 1/4 ( -20.64 ) = -5.16 ⁰F
Hence we can conclude that The temperature at which the Fahrenheit scale reads quartered the celsius scale is : -5.16 ⁰F, -20.64⁰C. Reading in kelvin scale = 252.5 K
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An electric current is characterized by which of the following?
a. Charges that are stationary
b. Charges that move
c. Potential energy in a battery
d. Mechanical energy in a system
Help pls
Answer:
B. Charges that move.
do most of the physical digestion take place in the top half of the digestive tract (mouth esophagus,and stomach ) or the bottom half (the intestine)?
A gas is contained in a chamber such as that in Figure 15.4. Suppose that the region outside the chamber is evacuated and the total mass of the block and the movable piston is 135 kg. When 2050 J of heat fl ows into the gas, the internal energy of the gas increases by 1730 J. What is the distance s through which the piston rises?
0.24 m distance through which the piston rises.
we have given the mass of piston+block = 135kg
2050 joule of heat flows into the gas
now;
According to first law thermodynamics,
dQ=du+dw
we have
2050=1730+MgS
Here M is mass, g is gravity, S is distance
now puttin gthe formula
MgS=320
S=320/1323
S= 0.24m
The link between the heat and work produced and the system's internal energy is explained by the first law of thermodynamics. The whole amount of energy present in a system, including the kinetic energy of molecules and the energy held inside their chemical bonds, is referred to as internal energy. The first law of thermodynamics states that the heat transferred into the system less the work done by the system equals the change in internal energy of the system. The law explains how energy is conserved when thermodynamic processes take place between two or more equilibrium states.To know more about first law of thermodynamics visit : https://brainly.com/question/3808473
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Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass 0.321 kg are dropped from a height of 2.15 m , but one of the balls is positively charged with q1 = 360 μC , and the second is negatively charged with q2=-360 μC .
Use conservation of energy to determine the difference in the speeds of the two balls when they hit the ground. (Neglect air resistance.)
v+−v− =
The difference in the speeds of the two balls when they hit the ground is mathematically given as
V=0.10m/s
What is the difference in the speeds of the two balls when they hit the ground.?Generally, the equation for Acceleration due to electric field is mathematically given as
[tex]A=g+\frac{q1E}{m}\\\\Therefore\\\\A=9.8*\frac{600*1000*150*}{0.520}\\\\A=9.97m/s2[/tex]
For Negative charge
[tex]A=g+\frac{q1E}{m}\\\\A=9.8*\frac{-600*1000*150*}{0.520}\\\\A=9.62m/s2[/tex]
In conclusion, the speed of the negative ball
[tex]v1=\sqrt{2ah}\\\\v1=\sqrt{2*9.67*2m}[/tex]
v1=6.21m/s
The difference in v
V=v1-v2
V=6.31-6.21
V=0.10m/s
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with full explaniation
A. The ball's (vertical) velocity [tex]v[/tex] at time [tex]t[/tex] is
[tex]v(t) = 30\dfrac{\rm m}{\rm s} - gt[/tex]
so that after 4 seconds, the ball's speed is
[tex]|v(4\,\mathrm s)| = \left|30\dfrac{\rm m}{\rm s} - \left(10\dfrac{\rm m}{\mathrm s^2}\right) (4\,\mathrm s)\right| = \boxed{10\frac{\rm m}{\rm s}}[/tex]
(The velocity is -10 m/s, so the ball is falling back down at this point.)
B. At maximum height, the ball has zero velocity, so it takes
[tex]30\dfrac{\rm m}{\rm s} - gt = 0 \implies t = \dfrac{30\frac{\rm m}{\rm s}}g = \boxed{3\,\mathrm s}[/tex]
for the ball to reach this height.
C. The height of the ball [tex]y[/tex] at time [tex]t[/tex] is
[tex]y(t) = \left(30\dfrac{\rm m}{\rm s}\right) t - \dfrac g2 t^2[/tex]
The maximum height is attained by the ball at 3 seconds after it's thrown, so
[tex]y_{\rm max} = \left(30\dfrac{\rm m}{\rm s}\right) (3\,\mathrm s) - \dfrac{10\frac{\rm m}{\mathrm s^2}}2 (3\,\mathrm s)^2 = \boxed{45\,\mathrm m}[/tex]
D. The time it takes for the ball to reach its maximum height is half the time it spends in the air. So the total airtime is [tex]\boxed{6\,\mathrm s}[/tex].
Put another way: When the ball returns to the height from which it was thrown, its final velocity has the same magnitude as its initial velocity but points in the opposite direction. This is to say, after the total time the ball is in the air, it's final velocity will be -30 m/s. Then the total airtime is
[tex]30\dfrac{\rm m}{\rm s} - gt = -30\dfrac{\rm m}{\rm s} \implies t = \dfrac{60\frac{\rm m}{\rm s}}g = \boxed{6\,\mathrm s}[/tex]
Put yet another way: Solve [tex]y(t) = 0[/tex] for [tex]t[/tex]. I don't see a need to elaborate...