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notsponge [240]

3 years ago

8

The wind exerts a force of 425 N to the north on a sailboat, while the water exerts a drag force of 325N in the opposite directi

on. What is the acceleration of the sailboat if the mass is 4100 kg?

1 answer:

Nataly_w [17]3 years ago

8 0

Answer:

its n vv

Explanation:

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What happens when the objects submerged in a fluid at rest?

Talja [164]

It will act upon a buoyant force on the magnitude of which is equal to weight of the fluid

3 0

3 years ago

A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l

Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>

Consider a body of mass m kept on a weighing machine in a lift.
The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
The reaction we get as the weight recorded by the machine, and it is called the apparent weight.

<h3>How to solve the question?</h3>

Here we have given with the actual weight of the body as 100lbs.
This 100lb child is standing on the scale or the weighing machine, when it is riding .
During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
There is also<em> mg </em>downwards and a normal reaction in the upward direction.
when we equate both the upward force and downward force, we get,

$ma=mg-N\\N=mg-ma$ i.e. during riding the scale reads a weight less than that of actual weight.

When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

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7 0

2 years ago

Which of the following statements are true for magnetic force acting on a current-carrying wire in a uniform magnetic field? Che

qaws [65]

Answer:

The following statements are correct.

1. The magnetic force on the current-carrying wire is strongest when the current is perpendicular to the magnetic field lines.

2. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the field.

3. The direction of the magnetic force acting on a current-carrying wire in a uniform magnetic field is perpendicular to the direction of the current.

Wrong statements:

1. The magnetic force on the current-carrying wire is strongest when the current is parallel to the magnetic field lines.

Explanation:

6 0

3 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force f is applied to the other end of th

Vlada [557]

For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

$F\times L = mg \times \frac{L}{2}$

here weight will act at mid point of door so its distance is half of the total distance where force is applied

here we know that

$mg = 145 N$

now we will have

$F = \frac{mg}{2}$

$F = \frac{145}{2} = 72.5 N$

so our applied force is 72.5 N

7 0

4 years ago

What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)

Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity $g=10m/sec^2$

And final value of acceleration due to gravity $g=9.8m/sec^2$

We have to find the percentage error

We know that percentage error is given by $percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100$

So $percentage\ error=\frac{10-9.8}{10}\times 100=2$ %

4 0

3 years ago