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nataly862011 [7]
3 years ago
13

A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal

l that brings him to rest. Assuming that he doesn't lose any speed during the fall or while sliding across the ice, how much work is done by the wall while stopping the ice skater?
Physics
1 answer:
sukhopar [10]3 years ago
3 0

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

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Mrrafil [7]

Answer:

a) Then net force acting on the wagon  30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

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The weight ( P = m*g =  10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force  Fn = P = 98 [N]  positive ( surface is frictionless)

∑Fy = 0     P - Fn = 0     P = Fn  = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N]  = 10 Kg * a                      [N] = Kg * m /s²

-30 [kg*m/s²]  = 10 * Kg * a

-30/10  =  - 3  m/s²  = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon  30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

5 0
3 years ago
A 530-g squirrel with a surface area of 935 cm2 falls from a 4.4-m tree to the ground. Estimate its terminal velocity. (Use the
Agata [3.3K]

Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=\frac{530}{1000}=0.530 kg

1kg=1000 g

Area=A=935 cm^2=935\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Height,h=4.4 m

C=1

\rho=1.21 kg/m^3

Width of rectangular prism,b=11.6 cm=\frac{11.6}{100}=0.116 m

1 m=100 cm

Length,l=23.2 cm=0.232 m

Area=l\times b=0.116\times 0.232=0.0269 m^2

Terminal velocity,v_t=\sqrt{\frac{2mg}{\rho CA}}

Where g=9.8 m/s^2

Using the formula

v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}

v_t=17.86 m/s

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Using the formula

v=\sqrt{2\times 9.8\times 4.4}

v=9.29 m/s

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2 years ago
Approximately how much air is in a column 1-cm2 in cross section that extends from sea level to the top of the atmosphere?
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Approximately 101 N air is in a column 1-cm2 in cross-section that extends from sea level to the top of the atmosphere

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At the equator, the radius of the Earth at sea level is 6378.137 km (3963.191 mi). At the poles, it is 6,356.752 km (3,949.903 km), and on average, it is 6,371.001 km (3,958.756 mi). The elevation of the shoreline—the boundary between the ocean and the land—is referred to as sea level. Land that is higher than this altitude is above sea level, and land that is lower is below sea level.

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Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

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= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

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= .588 x 1 Nm

Balancing these two torques we have

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