Ask question

Ask question

All categories

2 years ago

15

A 2.50 × 105 W motor is used for 26.4 s to pull a boat straight toward shore. How far does the boat move toward shore if a force

of 4.20 × 104 N is applied by the motor?

1 answer:

ozzi2 years ago

6 0

The power of the motor is
2.50x10⁵ W
or
2.50x10⁵ N m/s

The time it takes to pull the boat is
26.4 s

The force applied is
4.20x10⁴ N

The distance traveled by the boat is
2.50x10⁵ N m /s (26.4 s) / (4.20x10⁴ N)
= 157.14 m<span />

You might be interested in

Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

2 years ago

A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or

Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

$I_1= mr_1^2$

$I_2= mr_2^2$

since, $r_2^2$

$I_2^2$

therefore,

$\omega_1^2$

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0

2 years ago

A balance accurate to one-hundredth of a gram measures the mass of a rock to be 56.10 grams. How many significant digits are in

natali 33 [55]

There are 3 significant didgits

3 0

2 years ago

Read 2 more answers

The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth.

Talja [164]

Answer:

$F_g=461lb_f$

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

$V = \frac{4}{3} \pi r^3$

$V = 1.124*10^{9}ft^3$

So

$\delta_{iron}=m/V=491lb/ft^3$

$m=V*\delta=5.519*10^{11}lb$

Once you know both masses, you can calculate the force using Newton's universal law of gravitation:

$F_g=G\frac{m_1m_2}{d^2}$

Where G is the gravitational constant:

$G= 1.068846 * 10^{-9} ft^3 lb^{-1} s^{-2}$

$F_g=461lb_f$

4 0

2 years ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Lisa [10]

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

<u>Explanation:</u>

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

<u>A. Student 4: 9.61 m/s2 </u>

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 1.738%$ %.

<u>B. Student 3: 9.88 m/s2 </u>

Percentage of error = $\frac{9.88-9.78}{9.78} (100) = 1.022%$ %.

<u>C. Student 2: 9.79 m/s2 </u>

Percentage of error = $\frac{9.79-9.78}{9.78} (100) = 0.1022%$ % .

<u>D. Student 1: 9.78 m/s2</u>

Percentage of error = $\frac{9.78-9.61}{9.78} (100) = 0%$ % .

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

4 0

2 years ago