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labwork [276]
3 years ago
15

A piano of mass 852 kg is lifted to a height of 3.5 m. How much gravitational potential energy is added to the piano? Accelerati

on due to gravity is g = 9.8 m/s2 O A. 102,282 J O B. 29,224J O C. 2982 J O D. 304.3J SUSM​
Physics
1 answer:
slega [8]3 years ago
6 0

Answer:

29223.6J

Explanation:

Given parameters:

Mass of Piano = 852kg

Height of lifting  = 3.5m

Unknown:

Gravitational potential energy  = ?

Solution:

The gravitational potential energy of a body can be expressed as the energy due to the position of a body;

  G.P.E  = mgh

m is the mass

g is the acceleration due to gravity

h is the height

 Now insert the given parameters and solve;

   G.P.E  = 852 x 9.8 x 3.5 = 29223.6J

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Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on
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Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N

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3 years ago
Read 2 more answers
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
alekssr [168]

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:

        x_{f} = v_{ox} * t = v_{o} * cos \theta * t  (2)

  • 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:

        cos \theta = \frac{x_{f} }{v*t} = \frac{41.0m}{26.0m/s*3.00s} = 0.526 (3)

  • ⇒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:

       \Delta y = v_{oy} * t - \frac{1}{2} *g*t^{2} (6)

  • 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:

       \Delta y = 22.1 m/s* 3.0s - \frac{1}{2} *9.8m/s2*(3.00s)^{2} = 22.2 m (8)

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/s
  • The vertical component can be found applying the definition of acceleration (g in this case), solving for the final velocity, as follows:

       v_{fy} = v_{oy} - g*t  (9)

  • Replacing by the time t (a given), g, and  v₀y from (7), we can solve (9) as follows:

       v_{fy} = 22.1 m/s - 9.8m/s2*3.00s = -7.3 m/s  (10)

  • 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:

       v_{f} = \sqrt{(13.7m/s)^{2} +(-7.3m/s)^{2}} = 15.5 m/s (11)

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Unit 5 activity 2 I need answer
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Answer:

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A cart of mass 0.8 kg on a horizontal frictionless track has a tension of 4 N applied to on side of the cart. Calculate the acce
sweet [91]

Answer:

Option (c).

Explanation:

Given that,

Mass of a cart, m = 0.8 kg

Tension in the car, F = 4 N

Net force, F = ma

a is acceleration of the cart.

So,

a=\dfrac{F}{m}\\\\a=\dfrac{4\ N}{0.8\ kg}\\\\a=5\ m/s^2

So, the acceleration of the cart is 5\ m/s^2.

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4 years ago
If an open pipe of length vibrates in fundamental mode, then pressure variation is maximum
KIM [24]

Answer:

(B) at the middle of the pipe

Explanation:

In the case of an open pipe which vibrates in fundamental mode, an anti-node is formed at the middle of the pipe, here the amplitude of the wave is maximum. Hence, the pressure variation is also maximum at the middle.

8 0
3 years ago
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