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3 years ago

Kindly answer the question about Work and Power. Image is attached below.

Physics

1 answer:

natita [175]3 years ago

4 0

Answer:

89600 Watts.

Explanation:

From the question given above, the following data were obtained:

Mass (m) of speedboat = 2800 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Power (P) expended by the speedboat =?

Next, we shall determine the acceleration of the speedboat. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 16 m/s

Time (t) = 8 s

Acceleration (a) =?

v = u + at

16 = 0 + (a × 8)

16 = 0 + 8a

16 = 8a

Divide both side by 8

a = 16/8

a = 2 m/s²

The acceleration of the speedboat is 2 m/s²

Next, we shall determine the force exerted by the speedboat. This can be obtained as follow:

Mass (m) of speedboat = 2800 Kg

Acceleration (a) of speedboat = 2 m/s²

Force (F) of speedboat =?

F = ma

F = 2800 × 2

F = 5600 N

Finally, we shall determine the expended by the speedboat at maximum speed. This can be obtained as follow:

Force (F) of speedboat = 5600 N

Maximum velocity (v) = 16 m/s

Power (P) expended by the speedboat =?

P = F × v

P = 5600 × 16

P = 89600 Watts

Thus, the power expended by the speedboat is 89600 Watts.

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The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum

NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars $g_{m}=3.7\ m/s^2$

Time Period of pendulum on earth $T=1\ s$

Time period of Pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

for earth

$1=2\pi\cdot \sqrt{\frac{L}{9.8}}$

$L=\frac{9.8}{(2\pi )^2}$

$L=0.498\ m$

(b)For same time period on mars length is given by

$L'=\frac{g_m}{(2\pi )^2}$

$L'=\frac{3.7}{39.48}$

$L'=0.0936\ m$

$L'=9.36\ cm$

3 0

3 years ago

A particle released during the fission of uranium-235 is a(n) ________

Kaylis [27]

<span>A particle released during the fission of uranium-235 is a "Neutron"</span>

6 0

3 years ago

A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the

krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒ Ycm = (ms*D + mp*0) / (ms + mp)

⇒ Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0

3 years ago

If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo

Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0

3 years ago

A stationary bomb explodes in space breaking into a number of small fragments. At the location of the explosion, the net force d

Kaylis [27]

Answer:

.D)The Vector sum of the linear momenta of the fragments must be zero.

Explanation:

.D)The Vector sum of the linear momenta of the fragments must be zero.

This statement is true. This is so because no external force is acting on the masses. The motion is created by internal force so momentum of fragments will be conserved.

A) this statement is false because kinetic energy was zero in the beginning ( the bomb was stationary in the beginning )

B ) This statement is false because it violates the law of conservation of momentum .( it does not violates only when all the fragments have equal mass )

C ) This statement is zero because kinetic energy is not a vector quantity so two kinetic energy when added can not sum up to zero.

6 0

3 years ago