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3 years ago

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An object of mass m moves horizontally, increasing in speed from 0 to v in a time t. The power necessary to accelerate the objec

t during this time period is:
a: mv^2t/2
b: mv^2/2
c: 2mv^2
d: v sqrt /m2t
e: mv^2/2t

1 answer:

lana66690 [7]3 years ago

8 0

Answer:

Option e is correct.

Explanation:

An object of mass m moves horizontally, increasing in speed from 0 to v in a time t.

We know that,

Work done = change in kinetic energy

$W=\dfrac{1}{2}m(v^2-u^2)$

Here, u = 0

So,

$W=\dfrac{1}{2}mv^2$

Also,

Power = work done/time

So,

$P=\dfrac{\dfrac{1}{2}mv^2}{t}\\\\=\dfrac{mv^2}{2t}$

So, the correct option is (e).

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We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0

3 years ago

5.<br>why does the pointer rotate?

morpeh [17]

Answer: find the answer in the explanation.

Explanation:

From the experiment set up in the diagram, the pointer is resting on the drinking straw while the rod is resting on the drinking straw.

When the rod is being heated through the bursen burner, there will be linear expansion in the rod. As the rod increases its length, this causes the drinking straw to roll and thereby causing the pointer to rotate.

The pointer therefore rotates because of the thermer expansion that happen in the rod due to the heat from the bursen burner.

8 0

3 years ago

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

3 years ago

1. If a model train car with a momentum of 12 kg•m/s collides with a 2 kg model train car that is not moving, what is the total

wariber [46]

Answer:

$P_{f} = 12 \ kg.m/s$

Explanation:

Given data:

Momentum of moving model train, $P_{1} = 12 \ kg.m/s$

Mass of the stationary model train, $m = 2 \ kg$

Initial speed of the stationary model train, $v = 0$

Assume there is no external force is acting on the given train system.

In this case, the total linear momentum of the trains would be conserved.

Let the final linear momentum of the trains be $P_{f}.$

Thus,

$P_{i} = P_{f}$

$P_{1} + P_{2} = P_{f}$

$P_{1} + mv = P_{f}$

$12 + 2 \times 0 = P_{f}$

$\Rightarrow \ P_{f} = 12 \ kg.m/s.$

7 0

3 years ago

An object moving at 36 km/h reduces its speed to 18 km/h in 2sec. If the mass of the object is 20 kg, find the average force act

laiz [17]

Answer:

-50 N

Explanation:

Givens:

V_i = 36 km/h

V_f = 18 km/h

t = 2 s

m = 20 kg

First we have to convert our km/h into m/s:

(36 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 10 m/s

(18 km*(1000 m/1 km)) / (60 min *(60 s/1 min)) = 5 m/s

a = (V_f - V_i)/t

a = (5 m/s - 10 m/s) / 2 s

a = -2.5 m/s^2

F = m(a)

F = 20 kg(-2.5 m/s^2)

F = -50 N

It's a negative force meaning its acting on it opposite its current direction of movement.

7 0

3 years ago