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3 years ago

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If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s

2

2 answers:

enot [183]3 years ago

7 0

Answer:

a

Explanation:

goblinko [34]3 years ago

5 0

Answer:

D. 6 m/s²

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

a = (60 m/s − 0 m/s) / 10 s

a = 6 m/s²

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The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

3 0

3 years ago

3. When two things are the same temperature:

JulijaS [17]

The answer is definitely C.) their molecules move at the same average speed

3 0

3 years ago

A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.

Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

$a=\dfrac{v-u}{t}$ .............(1)

Now, the second equation of motion is :

$s=ut+\dfrac{1}{2}at^2$

Put the value of a in above equation as :

$s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2$

$s=\dfrac{t(u+v)}{2}$

$u=\dfrac{2s}{t}-v$

$u=\dfrac{2\times 47}{8.6}-2.3$

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0

3 years ago

Transverse, surface, and longitudinal waves are all __________ waves because they __________.

algol [13]

Answer:

a. mechanical; require a medium to travel through

Explanation:

Longitudinal, transverse and surface waves are types of mechanical waves. For example, within the longitudinal waves are the sound waves, which needs a medium to propagate like the air. This is why sound does not travel in a vacuum.

And an example of a transverse wave is the waves that form in the water when a rock is thrown (ripples), these waves need a medium (the water) to propagate.

On the other hand, electromagnetic waves such as light waves do not need a medium to propagate, this is why we can see the light of distant stars because their light travels through the vacuum until it reaches us.

So, the answer is:

Transverse, surface, and longitudinal waves are all mechanical waves because they require a medium to travel through .

5 0

3 years ago

4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)

11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

3 0

3 years ago