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Ierofanga [76]
11 months ago
10

a projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. what

is the projectile's horizontal speed after it has fallen 25 meters? 22 m/s 35 m/s
Physics
2 answers:
STatiana [176]11 months ago
5 0

Answer: 35m/s

Explanation: answer from castle learning

QveST [7]11 months ago
3 0

A projectile is launched horizontally from a height of 65 meters with an initial horizontal speed of 35 meters pre second. projectile's horizontal speed after it has fallen 25 meters is 35 m/s.

<h3>what is speed ?</h3>

Speed is measured as the ratio of distance to the time in which the distance was covered. Speed is a scalar quantity as it has only direction and no magnitude.

The formula of speed is represented as s=d/t, Where, s is the speed in m.s-1, d is the distance traveled in m, t is the time taken in s

There are four types of speed such as Uniform speed, Variable speed, Average speed, Instantaneous speed

Uniform speed is an uniform speed when the object covers equal distance at equal time intervals, variable speed is defined as when the object covers a different distance at equal intervals of times.

Average speed is defined as the uniform speed the ratio of total distance travelled by an object to the total time taken by the object.

Instantaneous speed is defined as an object is moving with variable speed, then the speed at any instant of time is known as instantaneous speed.

For more details regarding speed, visit

brainly.com/question/13263542

#SPJ2

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Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

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Answer:

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Explanation:

From the question given above, the following data were obtained:

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Force is related to momentum and time according to the following formula:

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With the above formula, we can calculate the force the white car experience during the collision. This can be obtained as illustrated below:

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