As we all know the epic relation between distance, speed and time; we cam easily estimate the time in which an object can reach to the destination or target.

As here in this case, we know the distance of the target and the velocity of a bullet exerted from a rifle given as follows,

Distance of the Target from the rifle edge = 324.0 m

Velocity of bullet exerted from the rifle = 720 m/s

Since we know that,

$\text {velocity}=\frac{\text {distance}}{\text {time}}$

$\text {Time}=\frac{\text {distance}}{\text {velocity}}$

We can simply implement all the values in the formula and get the results i.e the time required by the bullet to hit the target. Since both the values are in S.I units measures, we don't need to change or convert any of them. Hence,

$\text { Time }=\frac{324}{75}=0.45 \text { seconds }$

Therefore, the bullet will hit the target in 0.45 seconds.

4 0

3 years ago

When energy changes form, energy is... A) lost. B) created. C) not lost or created.

devlian [24]

Answer:

Explanation:

Energy is neither created or destroyed, but it can change forms...

6 0

2 years ago

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A 2 kg marble moving at 4 mi./s collides into a 1 kg marble at rest. After collision, the 2 kg marble speed decreased to 2 mi./s

Klio2033 [76]

Answer:

$2\sqrt{6}$ $\frac{mi}{s}$

Explanation:

Assuming there is no waste of energy:

$K_{1} = K_{2}\\\frac{1}{2}m_{1}v_{1_{1}}^{2} + \frac{1}{2}m_{2}v_{2_{1}}^2 = \frac{1}{2}m_{1}v_{1_{2}}^{2} + \frac{1}{2}m_{2}v_{2_{2}}^2\\\\=> m_{1}v_{1_{1}}^{2} + m_{2}v_{2_{1}}^2 = m_{1}v_{1_{2}}^{2} + m_{2}v_{2_{2}}^2\\\\m_{1} = 2 kg, m_{2} = 1 kg, v_{1_{1}} = 4 \frac{mi}{s} , v_{2_{1}} = 0\\=> 32 = 8 + v_{2_{2}}^{2} => v_{2_{2}} = 2\sqrt{6} \frac{mi}{s}$

6 0

2 years ago

A force of F = 125 N is used to drag a crate 3.0 m across a floor. The force is directed at an angle upward from the crate so th

Bezzdna [24]

Answer:

Explanation:

The forward force to drag is 125N

Distance of 3m.

The distance is on the positive horizontal 3i+0j

The force is directed upward so that the horizontal and vertical component is

Fx=107N and Fy=65N

F=107i+65j

Work done is dot product of Force and distance

Then,

W=F.ds

W=(107i+65j).(3i+0j)

Note i.i=j.j=1, i.j=j.i=0

Therefore

W=107×3+65×0

W=321J

The workdone done on the box is 321J

8 0

3 years ago

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