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2 years ago

How would I find the average distance for this?

Physics

2 answers:

NemiM [27]2 years ago

4 0

I think you should just add all of the distances and divide it by each one of the times

Maslowich2 years ago

3 0

Answer:

Explanation:add them then divide them by 100 I think

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In Ancient Greece, athletes competing in the long jump used handheld weights called halteres to lengthen their jumps. You are a

katovenus [111]

The halter add the distance to the jump in meters is 0.55 m.

<h3>What is projectile?</h3>

When an object is thrown at an angle from the horizontal direction, the object is said to be in projectile motion. The object which follows the projectile motion is called the projectile.

The magnitude of velocity u =10.3 m/s, angle of jumping θ = 22.8 degrees.

Components of velocity in x and y direction are

Vx = 10.3 cos 22.8 = 9.5 m/s

Vy = 10.3 sin 22.8 = 4 m/s

Maximum Range of athlete achieved using halter is given by

R = u²sin2θ /g

where, u = initial velocity, θ is the angle of projection and g is the gravitational acceleration.

Substituting the values, we get

R = (10.3)² sin(2 x 22.8 °) / 2 x 9.81

R = 7.75m

At the peak of jump you throw two 5.5 kg masses horizontally behind you such that their velocity is zero in the ground's reference frame.

The momentum is conserved in this situation,

(M+2m)Vxo =MVx'

Vx' = (M+2m)/M x Vxo'

Change in x component of velocity ΔVx = Vx' -Vxo

Vxo = 2m/M x Vx

Vxo = 2 x 5.5 /78 x 9.5

Vxo = 1.34 s

Maximum height gained when final velocity is zero

Vy = 0 = Vyo -gt

time t = Vyo/g = 4/9.8 = 0.41s'

Increase in range by using of halters is

ΔR = ΔVx' x t

ΔR = 1.34 x 0.41

ΔR =0.55m

Thus, the halter add the distance to the jump in meters is 0.55 m.

Learn more about projectile.

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3 0

2 years ago

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it stri

Anni [7]

Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight sThe x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.16227766016838, since the stone is thrown down (and to the right).(b) the initial speed m/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed m/sangle °

5 0

3 years ago

WILL GIVE BRAINLIEST IF CORRECT!

Norma-Jean [14]

Answer:

10000N

Explanation:

Given parameters:

Mass of the car = 1000kg

Acceleration = 3m/s²

g = 10m/s²

Unknown:

Weight of the car = ?

Solution:

To solve this problem we must understand that weight is the vertical gravitational force that acts on a body.

Weight = mass x acceleration due to gravity

So;

Weight = 1000 x 10 = 10000N

5 0

3 years ago

This picture represents the electric field diagram between two particles with static charges. Do the two particles have the same

dexar [7]

Answers:

No, They will attract each other, B, and neither direction

Explanation:

Since the two already presented particles in the diagram represent both opposing charges due to the direction of the arrows (the arrows facing away from the particle shows a positive charge and the particles facing towards the particle show a negative charge), not only because of this but as the arrows between the particles show an attracting magnetic field, then it can be concluded that the particles will attract to each other and if another particle was introduced into the diagram of a positive charge, then it would attract to the negatively charged particle. If you have any questions or need further explanation, please comment below. E2021, have a great day.

7 0

3 years ago

Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma

AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

$m_1$ = Mass of first car = 120 kg

$m_2$ = Mass of second car

$u_1$ = Initial Velocity of first car = 14 m/s

$u_2$ = Initial Velocity of second car = 0 m/s

$v_1$ = Final Velocity of first car = -2 m/s

$v_2$ = Final Velocity of second car

For perfectly elastic collision

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}$

Applying in the next equation

$v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s$

$m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg$

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0

3 years ago