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Zepler [3.9K]
2 years ago
9

Two students try to move a heavy box. One pushes with the force of the 20N while the other pulls with a force of 30N in the same

direction. What is the work done by each boy after 10 seconds if the box can\t be moved? Show your equation.
Physics
1 answer:
hichkok12 [17]2 years ago
6 0

Answer:

Explanation:

Time is not part of the Work equation. That's the only conclusion that you can come to.

Work = Force * distance.

Not enough information is given to go any further. You don't have enough information to calculate the distance.

We don't know if the box can be moved or not. It says heavy. 50 N is really not very much.

I would guess that you are intended to answer that the box didn't move, but it's really hard to tell.

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A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo
Naddika [18.5K]

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

\Delta L = L_f - L_i

\Delta L = 21.5 - 11.9 = 9.6 cm

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

mg = kx

0.037 (9.81) = k(0.096)

k = 3.78 N/m

Now the period of oscillation of spring is given as

T = 2\pi \sqrt{\frac{m}{k}}

Now plug in all values in it

T = 2\pi \sqrt{\frac{0.037}{3.78}}

T = 0.62 s

5 0
3 years ago
A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
2 years ago
A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.
Hunter-Best [27]

Answer:

Correct answer: Third statement  P = 4900 W

Explanation:

Given:

m = 500 kg  the mass of the elevator

h = 10 m  reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

God is with you!!!

3 0
3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
A spiral spring has a length of 14 cm when a force of 4 N is hung on it. A force of 6 N extends
stira [4]

Answer:

bussy

Explanation:

shart

5 0
1 year ago
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