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Gemiola [76]
3 years ago
8

A 50 kg trapeze artist falls from a height of 10 m, and has a speed of 14 m/s just before hitting the safety net. How much time

is needed in order for them to stop if the maximum amount of force they can withstand is 500N?
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

t = 1.4 s

Explanation:

momentum of the trapeze just before hitting the net = 50 x 14 kg m/s

= 700 kg m/s

final momentum = 0

change in momentum required = 700 - 0 = 700

If F be the maximum force and t be the time of action of force

impulse = force x time = F x t

500 x t

impulse = change in momentum

500 x t = 700

t = 1.4 s .

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A 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the
Nesterboy [21]

Answer:

A.) 1372 N

B.) 1316 N

C.) 1428 N

Explanation:

Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:

a. The load moves downward at a constant velocity

At constant velocity, acceleration = 0

T - mg = ma

T - mg = 0

T = mg

T = 140 × 9.8

T = 1372N

b. The load accelerates downward at a rate 0.4 m/s??

Mg - T = ma

140 × 9.8 - T = 140 × 0.4

1372 - T = 56

-T = 56 - 1372

- T = - 1316

T = 1316N

C. The load accelerates upward at a rate 0.4 m/s??

T - mg = ma

T - 140 × 9.8 = 140 × 0.4

T - 1372 = 56

T = 56 + 1372

T = 1428N

8 0
3 years ago
HELP ME PLEASE AND HURRY !!!!!!
hodyreva [135]

Answer:

mass number = protons + neutrons

8 0
2 years ago
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The record distance in the sport of throwing cowpats is 81.1 m. This record toss was set by Steve Urner of the United States in
N76 [4]

Answer:

28.2 m/s

Explanation:

The range of a projectile launched from the ground is given by:

d=\frac{v^2}{g}sin 2\theta

where

v is the initial speed

g = 9.8 m/s^2 is the acceleration of gravity

\theta is the angle at which the projectile is thrown

In this problem we have

d = 81.1 m is the range

\theta=45^{\circ} is the angle

Solving for v, we find the speed of the projectile:

v=\sqrt{\frac{dg}{sin 2 \theta}}=\sqrt{\frac{(81.1 m)(9.8 m/s^2)}{sin (2\cdot 45^{\circ})}}=28.2 m/s

7 0
3 years ago
Air at 20ºC with a convection heat transfer coefficient of 20 W/m2·K blows over a pond. The surface temperature of the pond is 2
dimulka [17.4K]

Answer:

The heat flux between the surface of the pond and the surrounding air  is<em> 60 W/</em>m^{2}<em> </em>

Explanation:

Heat flux is the rate at which heat energy moves across a surface, it is the heat transferred per unit area of the surface. This can be calculated using the expression in equation 1;

q = Q/A ...............................1

since we are working with the convectional heat transfer coefficient equation 1 become;

q = h (T_{sf} -T_{sd}) ........................2

where q is the  heat flux;

Q is the heat energy that will be transferred;

h is the convectional heat coefficient = 20 W/m^{2}.K;

T_{sf} is the surface temperature = 23^{o}C 23°C + 273.15  = 296.15 K;

T_{sd} is the surrounding temperature = 20^{o}C = 20°C + 273.15  = 293.15 K;

The values are substituted into equation 2;

q = 20 W/m^{2}.K (  296.15 K - 293.15 K)

q = 20 W/m^{2}.K ( 3 K)

q =  60 W/m^{2}

Therefore the heat flux between the surface of the pond and the surrounding air  is 60 W/m^{2}

3 0
4 years ago
You are on the roof of a building, 46.0 m above the ground. Your friend, who is 1.80 m tall, is standing next to the building. W
ExtremeBDS [4]

Answer:

a.) 866 m/s

beacues

5 0
3 years ago
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