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skelet666 [1.2K]
2 years ago
10

URGENT! 40 POINTS! For each situation below, find the work done on the box using both energy and forces.

Physics
1 answer:
TiliK225 [7]2 years ago
4 0

Answer:

i’m not sure one second let me try to figure it out

Explanation:

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You can enter compound units that are combinations of other units that are multiplied together. to enter the ⋅ explicitly, type
alexdok [17]

Answer:

30 N \cdot m

Explanation:

The torque applied by a force can be calculated as

\tau = F d sin \theta

where

F is the magnitude of the force

d is the length of the arm

\theta is the angle between the direction of the force and the arm

In this problem, we have

F = 15 N

d = 2.0 m

\theta=90^{\circ}

Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

7 0
3 years ago
A roller coaster is stationary at the top of a ramp.
Serjik [45]

Answer:

A

Explanation:

The roller coaster is stationary so the kinetic energy would be zero, but it is at the top of ramp so the potential energy would be high as its gravitational  so it would have to be A

6 0
2 years ago
A centrifuge in a forensics laboratory rotates at an angular speed of 3,700 rev/min. When switched off, it rotates 46.0 times be
ra1l [238]

Answer:

Explanation:

Given,

initial angular speed, ω = 3,700 rev/min

                                      = 3700\times \dfrac{2\pi}{60}=387.27\ rad/s

final angular speed = 0 rad/s

Number of time it rotates= 46 times

angular displacement, θ = 2π x 46 = 92 π

Angular acceleration

\alpha = \dfrac{\omega_f^2 - \omega^2}{2\theta}

\alpha = \dfrac{0 - 387.27^2}{2\times 92\ pi}

\alpha = -259.28 rad/s^2

3 0
3 years ago
Suppose you ran 53 km in 67 min. With what speed did you run?
melomori [17]

Answer:

47.46 kilometers an hour

Explanation:

hope this helped

4 0
2 years ago
Read 2 more answers
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
2 years ago
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