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4 years ago

13

If a boy (m = 50kg) at rest on skates is pushed by another boy who exerts a force of 200 N on him and if the first boy's final v

elocity is 8 m/s, what was the contact time?

2 answers:

AVprozaik [17]4 years ago

8 0

<span>F* t = (m x v_final) - (m x v_initial)

200 x t = 50x8 - 50x0

t= 50 x 8 /200

t= 2s </span>

andrew11 [14]4 years ago

8 0

The answer would be 2 seconds

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Seema knows the mass of basketball. What other information is needed to find the balls potential energy

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Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field. In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy $U$ will be:

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Where:

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$h$ is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

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3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

bogdanovich [222]

Answer:

Toyota Camry

F_d = 51.852 N

F_d = 100.042 N

Hummer H2

F_d = 412.0888 N

F_d = 8351.755 N

Explanation:

Given:

- The density of air p_air = 1.2 kg/m^3

- The drag force equals car's engine force.

Find:

- What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.)

- What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.)

Solution:

- The formula for drag force is given as follows:

F_d = 0.5*C_d*p_air*A*V^2

Where,

A : The drag Area m^2

C_d: The drag coefficient

V: Velocity m/s

a) Toyota Camry

C_d = 0.28 , A = 0.70 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

F_d = 0.5*0.28*1.2*0.70*(22.22)^2

F_d = 51.852 N

C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

F_d = 0.5*0.28*1.2*0.70*(29.1667)^2

F_d = 100.042 N

b) Hummer H2

C_d = 0.57 , A = 2.44 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

F_d = 0.5*0.57*1.2*2.44*(22.22)^2

F_d = 412.0888 N

C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

F_d = 0.5*0.57*1.2*2.44*(29.1667)^2

F_d = 8351.755 N

4 0

3 years ago

an object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34

Rzqust [24]

-244.59meters

Explanation:

The initial position of the object is located at -244.59meters from the origin.

Displacement is the distance from the start to finish of a body.

Final position of the body is at x=-55.25 meters

Amount of displacement is -189.34meters;

let us represent this by a coordinate system in the negative x - direction;

Origin

Start finish 0m

-a -x

-55.25meters

displacement

-189.34meters

Position -a is about 189.34 in left direction from 55.25; this gives: 55.25 + 189.34 = -244.59m in the left direction

learn more:

Displacement brainly.com/question/5461768

#learnwithBrainly

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3 years ago