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2 years ago

In the graph above, what is the instantaneous speed of the object after the first five seconds?

Physics

1 answer:

alexdok [17]2 years ago

8 0

Answer:

D) 25 m/s

Explanation:

In order to solve this problem we must use the following kinematics equation.

$v_{f} =v_{i} +(a*t)\\$

where:

Vf = final speed [m/s]

Vi = initial speed = 0

a = acceleration = 5[m/s^2]

t = time = 5[s]

After 5 seconds the acceleration is equal to 5 [m/s^2]

Now replacing the values in the equation:

Vf = 0 + (5*5)

Vf = 25[m/s]

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MissTica

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2 years ago

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/

PilotLPTM [1.2K]

Answer:

a) Vf = 27.13 m/s

b) It would have been the same

Explanation:

On the y-axis:

$Y=-Vo*sin\theta*t-1/2*g*t^2$

$-8=-24*sin(21)*t-1/2*10*t^2$

Solving for t:

t1 = 0.67s t2= -2.4s

Discarding the negative value and using the positive one to calculate the velocity:

$Vf_y = -Vo*sin\theta-g*t$

$Vf_y = -15.3m/s$

So, the module of the velocity will be:

$Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}$

$Vf=27.13m/s$

If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.

3 0

3 years ago

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

Firlakuza [10]

Answer:

$v_B=\frac{v}{3}$

Explanation:

Given that:

mass of object A, $m_A=M$

mass of object B, $m_B=2M$

speed of object A, $v_A=v$

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

$m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B$

$M\times v+0 = M\times \frac{v}{3}+2M\times v_B$

$2M\times v_B= \frac{2M\times v}{3}$

$v_B=\frac{v}{3}$

3 0

3 years ago

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Vera_Pavlovna [14]

Answer:1.14.102 on the side of 10

Explanation:

6 0

3 years ago

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dlinn [17]

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