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Scilla [17]
3 years ago
14

Which has a total mass of 1612 kg. If she accelerates from rest to a speed of 12.87 m/s in 3.47 s, what is the minimum power req

uired of the engine?
Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

76969.29 W

Explanation:

Applying,

P = F×v............. Equation 1

Where P = Power, F = force, v = velocity

But,

F = ma.......... Equation 2

Where m = mass, a = acceleration

Also,

a = (v-u)/t......... Equation 3

Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s

Substitute these values into equation 3

a = (12.87-0)/3.47

a = 3.71 m/s²

Also Given: m = 1612 kg

Substitute into equation 2

F = 1612(3.71)

F = 5980.52 N.

Finally,

Substitute into equation 1

P = 5980.52×12.87

P = 76969.29 W

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The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters
skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

<h2>ΔT = 17.11 °C</h2>

Hope this helps

5 0
3 years ago
A piece of wood has a volume of 4.6 cm to the power of 3, and a mass of 3.7 g what is the density of the wood
AysviL [449]

Density is given by:

D = M/V

D = density, M = mass, V = volume

Given values:

M = 3.7g, V = 4.6cm³

Plug in and solve for D:

D = 3.7/4.6

D = 0.80g/cm³

3 0
3 years ago
Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity  because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion



5 0
4 years ago
a radio antenna broadcasts a 1.0 MHz radio wav e with 21 kW of power. Assume that the radiation is emitted uniformly in all dire
My name is Ann [436]

Answer:

I=2.67\times 10^{-6}\ W/m

Explanation:

Given that,

Frequency of a radio antenna is 1 MHz

Power, P = 21 kW

We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :

I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{21\times 10^3}{4\pi (25000)^2}\\\\I=2.67\times 10^{-6}\ W/m

So, the wave intensity 25 km from the antenna is 2.67\times 10^{-6}\ W/m.

5 0
3 years ago
6
hodyreva [135]

Answer:....

Explanation:.

6 0
3 years ago
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