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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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Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

Kinetic energy = (1/2)mv²

$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

$Vf = \sqrt{frac{22.4}{2}$

= 3.33 m/s

b) magnitude of force when x = 5.0m

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

8 0

3 years ago

(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i

murzikaleks [220]

Answer:

a) $t = -\frac{ln(2)}{k}$

b) See the proof below

$A(t) = A_o 2^{-\frac{t}{T}}$

c) $t = 3T \frac{ln(2)}{ln(2)}= 3T$

Explanation:

Part a

For this case we have the following differential equation:

$\frac{dA}{dt}= kA$

With the initial condition $A(0) = A_o$

We can rewrite the differential equation like this:

$\frac{dA}{A} =k dt$

And if we integrate both sides we got:

$ln |A|= kt + c_1$

Where $c_1$ is a constant. If we apply exponential for both sides we got:

$A = e^{kt} e^c = C e^{kt}$

Using the initial condition $A(0) = A_o$ we got:

$A_o = C$

So then our solution for the differential equation is given by:

$A(t) = A_o e^{kt}$

For the half life we know that we need to find the value of t for where we have $A(t) = \frac{1}{2} A_o$ if we use this condition we have:

$\frac{1}{2} A_o = A_o e^{kt}$

$\frac{1}{2} = e^{kt}$

Applying natural log we have this:

$ln (\frac{1}{2}) = kt$

And then the value of t would be:

$t = \frac{ln (1/2)}{k}$

And using the fact that $ln(1/2) = -ln(2)$ we have this:

$t = -\frac{ln(2)}{k}$

Part b

For this case we need to show that the solution on part a can be written as:

$A(t) = A_o 2^{-t/T}$

For this case we have the following model:

$A(t) = A_o e^{kt}$

If we replace the value of k obtained from part a we got:

$k = -\frac{ln(2)}{T}$

$A(t) = A_o e^{-\frac{ln(2)}{T} t}$

And we can rewrite this expression like this:

$A(t) = A_o e^{ln(2) (-\frac{t}{T})}$

And we can cancel the exponential with the natural log and we have this:

$A(t) = A_o 2^{-\frac{t}{T}}$

Part c

For this case we want to find the value of t when we have remaining $\frac{A_o}{8}$

So we can use the following equation:

$\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}$

Simplifying we got:

$\frac{1}{8} = 2^{-\frac{t}{T}}$

We can apply natural log on both sides and we got:

$ln(\frac{1}{8}) = -\frac{t}{T} ln(2)$

And if we solve for t we got:

$t = T \frac{ln(8)}{ln(2)}$

We can rewrite this expression like this:

$t = T \frac{ln(2^3)}{ln(2)}$

Using properties of natural logs we got:

$t = 3T \frac{ln(2)}{ln(2)}= 3T$

8 0

3 years ago

An object has a momentum of 4,000 kg-m/s and a mass of 115 kg. It crashes into another object that has a mass of 100 kg, and the

ANTONII [103]

Answer:

18.60 m/s

Explanation:

Original momentum = mv = 4000 with m = 115

after collision m = 115 + 100 = 215 kg

but the total momentum is still the same (conserved)

4000 = 215 v shows v = 18.60 m/s

4 0

2 years ago

Define parallax as it relates to stars and explain how it can be used to determine stellar distances.

FrozenT [24]

Astronomers can measure a star's position once, and then again 6 months later and calculate the apparent change in position. The star's apparent motion is called stellar parallax. The distance d is measured in parsecs and the parallax angle p is measured in arcseconds.

I hope this helps!

5 0

3 years ago

A mass of 10kg is a point a on a table is moved to a point B.if the line joining A and is horizantal what is the done on the obj

IrinaK [193]

Answer:

work done on the object by gravitational force = 0 joules

Explanation:

work = force × displacement

force = mass × acceleration

so,

work = mass × acceleration × displacement

we know that mass= 10 kg , gravitational acceleration= 9.8 $m/s^2$

and displacement= 0 m since the object is not moving vertically.

so,

work = 10 × 9.8 × 0 = 0 joules

6 0

2 years ago

Read 2 more answers