Answer:
13.4 m/sExplanation:
given
Mass of cart= 150kg
mass of each wheel=45kg
mass of 4 wheels= 180kg
angle of the track= 17 ∘
distance of track= 27m
The height of the tracl is calculated thus:
sin 17° = h / 27
h = sin 17*27
h=7.89m
"Potential energy at top = kinetic energy of cart and wheels at bottom + rotational energy of wheels at bottom "
1. Potential energy at top= (M+4m)gh
2. kinetic energy of cart and wheels at bottom= 1/2 (M+4m) v²
3. rotational energy of wheels at bottom= 4(1/2 Iω²)
The total is expressed as
(M+4m)gh = 1/2 (M+4m) v² + 4(1/2 Iω²) -------------1
we know that I = mr² / 2
Put I= mr² / 2
(M+4m)gh = 1/2 (M+4m)v² + 4(1/2 (mr² / 2) ω²)
(M+4m)gh = 1/2 (M+4m)v² + m r² ω²
we know that v²= r² ω²
(M+4m)gh = 1/2 (M+4m)v² + m v²
(M+4m)gh = v² (M/2 + 2m + m)
(M+4m)gh = v² (M/2 + 3m)
v = √[(M+4m)gh / (3m + M/2)]
v = √[(150 + 4*45) * 9.8 m/s² * 7.89 m / (3*45 + 150/2)]
v=√25516.26/142.5
v=√179.06
v = 13.4 m/s
Average formula is...?
Answer:
total sum of all numbers/ number of items in the set
Explanation:
Calculate the force necessary to accelerate a 10 kg table from
O m/s to 4 m/s in 2 seconds.
Answer:
a= v/t
a = 4/2
a = 2 m/s^2
And F = M a
F = 10 × 2
F = 20 N
PLEASE HELP!!!!
On his quest to find a shrubbery for the "Knights Who Say Ni!", King Arthur rides his horse off a 15m cliff and lands 4.3m away.
What was his original horizontal velocity?
Answer:
v = 2.45 m/s
Explanation:
first we find the time taken during this motion by considering the vertical motion only and applying second equation of motion:
h = Vi t + (1/2)gt²
where,
h = height of cliff = 15 m
Vi = Initial Vertical Velocity = 0 m/s
t = time taken = ?
g = 9.8 m/s²
Therefore,
15 m = (0 m/s) t + (1/2)(9.8 m/s²)t²
t² = (15 m)/(4.9 m/s²)
t = √3.06 s²
t = 1.75 s
Now, we consider the horizontal motion. Since, we neglect air friction effects. Therefore, the horizontal motion has uniform velocity. Therefore,
s = vt
where,
s = horizontal distance covered = 4.3 m
v = original horizontal velocity = ?
Therefore,
4.3 m = v(1.75 s)
v = 4.3 m/1.75 s
v = 2.45 m/s
Is a man kicking ball potential or kinetic?
during a basketball game what are the most points you can score from shooting and getting fouled
Answer:
If a player is fouled while shooting a three-point shot and makes it anyway, he is awarded one free throw. Thus, he could score four points on the play. Inbounds- If fouled while not shooting, the ball is given to the team the foul was committed upon.
Is a light bulb that is on potential or kinetic?
Answer:
pretty sure its kinetic
Explanation:
Answer:
Kinetic
Explanation:
The stored chemical potential energy of a battery converts to electrical kinetic energy to transport electricity to a light bulb, which radiates thermal kinetic energy.
A car is filled up with 20 gallons of gas. The car uses .25 gallons per minute. How much time will the car travel ?
WILL MARK BRAINLIEST -- The Earth’s velocity is shown by a red vector. Which best describes the Earth’s velocity?
a. It is constant in magnitude and direction
b. It is constant in magnitude but not constant in direction
c. It is not constant in magnitude but is constant in direction
d. It is not constant in magnitude or direction
The chemists'_
is another name for the periodic table.
Answer here
Answer: Calendar
Explanation: I'm pretty sure it is calendar.
Some nitrogen at 80.0 psi gauge pressure occupies 13.0 ft^3. Find its volume, in ft^3, at 50.0 psi gauge pressure.
Answer:
20.8 ft³
Explanation:
The following data were obtained from the question:
Initial pressure (P1) = 80 psi
Initial volume (V1) = 13 ft³
Final pressure (P2) = 50 psi
Final volume (V2) =?
The new volume of the gas can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
80 × 13 = 50 × V2
1040 = 50 × V2
Divide both side by 50
V2 = 1040 / 50
V2 = 20.8 ft³
Thus, the volume of the gas at a pressure of 50 psi is 20.8 ft³
In a galvanic cell, electrons flow from the ________ to the _______ .
A. salt bride, anode
B. anode, salt bridge
C. cathode, anode
D. salt bridge, cathode
E. anode, cathode
Answer:
E.
Explanation:
In a galvanic cell, electrons flow from the anothe to the cathode.
I hope you got the answer
Why is science important for society? (2 points)
a
It helps make decisions based on scientific reasoning.
b
It proves that all questions can be answered using science.
c
It emphasizes the importance of human choice in decision making.
d
It shows that all social problems can be solved using scientific knowledge.
Answer:
i think its d
Explanation:
i took this quiz and got 100 percent
The mass of Earth is 5.972×1024 kg and its orbital radius is an average of 1.496×1011 m . Calculate its linear momentum.
A block is at rest. The coefficients of static and kinetic friction are s = 0.81 and k = 0.69, respectively. The acceleration of gravity is 9.8 m/s^2.
Required:
a. What is the largest angle which the incline can have so that the mass does not slide down the incline?
b. What is the acceleration of the block down the incline if the angle of the incline is 44°?
Answer:
Explanation:
a ) The angle required = angle of repose = θ
Tanθ = .81
θ = 39⁰
b ) when angle of incline θ = 44
Net force on the block = mg sinθ - μ mg cosθ where μ is coefficient of kinetic friction
acceleration = gsinθ - μ g cosθ
= 9.8 ( sin44 - μ cos44 )
= 9.8 ( .695 - .69 x .72 )
= 9.8 ( .695 - .497 )
= 1.94 m /s²
A particle is fallingdown into a medium with an initial velocity of 30m/s. If the acceleration of the particle is =(−4t)/m/s^2 , where t is in seconds, determine the distance traveled before the particle stops.
Answer:
The distance traveled by the particle before it stops is 41.06 m.
Explanation:
We can find the distance traveled by the particle using the following equation:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex] v_{f}[/tex]: is the final velocity = 0 (when it stops)
[tex] v_{0}[/tex]: is the initial velocity = 30 m/s
a: is the acceleration = -4t m/s²
t: is the time
d: is the distance
First, we need to calculate the time:
[tex] v_{f} = v_{0} + at [/tex]
[tex] 0 = 30 m/s + (-4t m/s^{2})t [/tex]
[tex]0 = 30 m/s - 4t^{2} m/s^{3}[/tex]
[tex]t = 2.74 s[/tex]
Now, the acceleration is:
[tex] a = -4t = -10.96 m/s^{2} [/tex]
Hence, the distance is:
[tex] d = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{-(30 m/s)^{2}}{2*(-10.96 m/s^{2})} = 41.06 m [/tex]
Therefore, the distance traveled by the particle before it stops is 41.06 m.
I hope it helps you!
What is the critical angle for light traveling from crown glass (n = 1.52) into water (n = 1.33)?
42 degrees
48 degrees
57 degrees
61 degrees
Answer:
61 degrees, I just did the test.
Explanation:
Answer: 61 degrees
Explanation:
I just did the question and got it right
Calculate the height from which a body is released rest if its velocity just before hitting the ground is 30ms-1
Answer:
45.9m
Explanation:
Given parameters:
Final velocity = 30m/s
Initial velocity = 0m/s
Unknown:
Height of fall = ?
Solution:
The motion equation to solve this problem is given below;
V² = U² + 2gH
V² = 0 + (2 x 9.8 x H)
30² = 19.6H
H = [tex]\frac{900}{19.6}[/tex] = 45.9m
A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If the merry-go-round makes 4.9 rev/min, what is the velocity of the child in m/s?
a.
9.2 m/s
b.
4.6 m/s
c.
0.74 m/s
d.
1.75 m/s
Answer:
b. 4.6 m/sExplanation:
the formula for calculating the linear velocity is expressed as;
v = wr
w is the angular velocity
r is the radius
r = d/2 = 18/2
r = 9 m
Given that 1rev/min = 0.10472rad/s
4.9 rev/min = x
x = 4.9 * 0.10472
x = 0.513128 rad/s
Substitute into the given formula;
v = 9 * 0.513128
V = 4.61m/s
Hence the velocity of the child in m/s is 4.6m/s
Is a ball standing still potential or kinetic?
A stone sphere of radius 7.00 m rests in a flat field. Relative to the ground, what is the gravitational potential energy of a 90.0-kg person sitting on the very top of the sphere?
a. 7.76 x 104 J
b. 1.23 x 104 J
c. 3.88 x 104 J
Answer:
(B) 1.23 x 10⁴ J
Explanation:
Given;
radius of the sphere, r = 7.0 m
diameter of the sphere, d = 2r = 14.0 m
mass of the person sitting on the sphere, m = 90.0 kg
The gravitational potential energy of the person is given by;
P.E = mgh
where;
g is acceleration due to gravity = 9.8 m/s²
h is the height above the ground level = d = 14.0 m
P.E = mgh
P.E = (90)(9.8)(14)
P.E = 12348 J
P.E = 1.2348 x 10⁴ J
Therefore, the gravitational potential energy of the person is 1.2348 x 10⁴ J
big ideas
key questions and terms notes
Answer:
what i and confused
Explanation:
Answer:
nerd
Explanation:
I H4+3 U
A laser (electromagnetic wave) has the maximum electric field strength of 1.0x1011 V/m. What is the force the laser applies on a mirror (totally reflective) of 5.0 mm2 area? A. 2.76 x105N B. 1.21 x106N C. 1.94 x106N D.4.43 x105 N E. 7.82 x104N
Answer:
The correct option is D
Explanation:
From the question we are told that
The maximum electric field strength is [tex]E = 1.0 *10^{11} \ V/m[/tex]
The area is [tex]A = 5.0 \ mm^2 = 5.0 *10^{-6} \ m^2[/tex]
Generally the force the laser applies is mathematically represented as
[tex]F = \epsilon_o * E ^2 * A[/tex]
Here [tex]\epsilon_o = 8.85*10^{-12} C/(V \cdot m)[/tex]
[tex]F = 8.85*10^{-12} * (1.0 *10^{11}) ^2 * 5.00*10^{-6 }[/tex]
=> [tex]F = 4.43 *10^{5} \ N[/tex]
How did the atomic model changed from daltons model in 1803 to schrodingers model of 1926? In 3 to 4 sentences, describe the changes and explain whether the discoveries were independent or cited previous discoveries.
Explanation:
Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles .
Suppose we wish to use a 8.0 m iron bar to lift a heavy object by using it as a lever. If we place the pivot point at a distance of 1.0 m from the end of the bar that is in contact with the load and we can exert a downward force of 562 N on the other end of the bar, find the maximum load that this person is able to lift (pry) using this arrangement (neglect the mass of the bar in this problem).
Answer:
W = 3934 N , m = 401.43 kg
Explanation:
This problem can be solved using the rotational equilibrium relation, where we place the reference system at the pivot point and assume that the counterclockwise turns are positive.
∑τ = 0
F 7 - W 1 = 0
W = F 7/1
W = 562 7
W = 3934 N
W = mg
m = W / g
m = 3934 / 9.8
m = 401.43 kg
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force of the earth were to act on it (ie., neglect the forces from the sun and other solar system objects), the spacecraft would eventually crash into the earth The mass of the earth is Me and its radius is Re. Neglect air resistance throughout this problem, since the spacecraft is primarily moving through the near vacuum of space
Find the speed s of the spacecraft when it crashes into the earth Express the speed in terms of M, Re, and the universal gravitational constant G.
Answer:
Speed of the spacecraft right before the collision: [tex]\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}[/tex].
Assumption: the earth is exactly spherical with a uniform density.
Explanation:
This question could be solved using the conservation of energy.
The mechanical energy of this spacecraft is the sum of:
the kinetic energy of this spacecraft, andthe (gravitational) potential energy of this spacecraft.Let [tex]m[/tex] denote the mass of this spacecraft. At a distance of [tex]R[/tex] from the center of the earth (with mass [tex]M_\text{e}[/tex]), the gravitational potential energy ([tex]\mathrm{GPE}[/tex]) of this spacecraft would be:
[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}[/tex].
Initially, [tex]R[/tex] (the denominator of this fraction) is infinitely large. Therefore, the initial value of [tex]\mathrm{GPE}[/tex] will be infinitely close to zero.
On the other hand, the question states that the initial kinetic energy ([tex]\rm KE[/tex]) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.
Right before the collision, the spacecraft would be very close to the surface of the earth. The distance [tex]R[/tex] between the spacecraft and the center of the earth would be approximately equal to [tex]R_\text{e}[/tex], the radius of the earth.
The [tex]\mathrm{GPE}[/tex] of the spacecraft at that moment would be:
[tex]\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex].
Subtract this value from zero to find the loss in the [tex]\rm GPE[/tex] of this spacecraft:
[tex]\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}[/tex]
Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the [tex]\rm GPE[/tex] of this spacecraft would be equal to the size of the gain in its [tex]\rm KE[/tex].
Therefore, right before collision, the [tex]\rm KE[/tex] of this spacecraft would be:
[tex]\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}[/tex].
On the other hand, let [tex]v[/tex] denote the speed of this spacecraft. The following equation that relates [tex]v\![/tex] and [tex]m[/tex] to [tex]\rm KE[/tex]:
[tex]\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2[/tex].
Rearrange this equation to find an equation for [tex]v[/tex]:
[tex]\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}[/tex].
It is already found that right before the collision, [tex]\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}[/tex]. Make use of this equation to find [tex]v[/tex] at that moment:
[tex]\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}[/tex].
The capacitors are reconnected in series, and the combination is again connected to the battery. From the same choices, choose the one that is true.
a) The potential difference across each capacitor is the same, and the equivalent capacitance is greater than any of the capacitors in the group.
b) The capacitor with the smallest capacitance carries the largest charge.
c) The potential differences across the capacitors are the same only if the capacitances are the same.
d) All capacitors have the same charge, and the equivalent capacitance is greater than the capacitance of any of the capacitors in the group.
e) The capacitor with the largest capacitance carries the smallest charge.
Answer:
This question appear incomplete
Explanation:
This question appear incomplete. However, when capacitors are connected in series, the total capacitance is usually less than the capacitance of individual capacitor. This is because the total capacitance can be calculated as 1 ÷ 1/C₁ + 1/C₂ +1/C₃...
The formula above does not refer to the charge; this is because capacitors connected in series have the same charge (regardless of the capacitance). Also, the largest potential difference occur in the capacitor with the smallest/lowest capacitance.
A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour her average speed is____ block/hour
Answer:
2 m/s
Explanation:
The total time = 1 hour
The vertical displacement = 1 - 1
Vertical displacement = 0
Horizontal displacement = 4 - 2
Horizontal displacement = 2
Total displacement = sqrt (2^2 - 0^2)
Displacement - 2
Average velocity is displacement/time
= 2x1
= 2 m/s
The average velocity is 2 metres per second.
A mass M is attached to a spring with spring constant k. When this system is set in motion with amplitude A, it has a period T. What is the period if the mass is doubled to 2M
Answer:
T' = o.707T
Explanation:
[tex]T = 1/2\pi \sqrt{k/m} \\\\m'=2m then\\T' = 1/2\pi \sqrt{k/m'}\\ = 1/2\pi \sqrt{k/2m}\\T' = \frac{(1/2\pi \sqrt{k/m}\\)}{1.4} T' = \frac{T}{1.4}\\T' = 0.707T[/tex]
Question #2
Like liquids, are gases made of particles?
Yes
No
Maybe
Answer:
Yes
Explanation:
Answer:
YesExplanation:
Remember that all solids, liquids, and gases are made up of atoms, molecules, and / or ions. ( Also remember that the answer "maybe" most likely won't be the answer in most quizzes. )
Which answers your question, gases are made up of particles.
Hope this helps! <3
7. Two capacitors, 5.8 µF and 2.1 µF, are connected in parallel. A 3 V DC voltage is applied across the capacitors. What would be the expected accumilative charge stored in the capacitors
Answe thanks for the freee points UwU
Explanation: