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8090 [49]
3 years ago
9

A 350-kg sailboat has an acceleration of 0.62 m/s2 at a angle of 64 degrees north of east find the magnitude and direction of th

e net force that acts on the sail boat
Physics
1 answer:
Kay [80]3 years ago
4 0

Answer:

<em>The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.</em>

Explanation:

<u>Mechanical Force</u>

The second Newton's law states the net force exerted by an external agent on an object of mass m is:

\vec F = m.\vec a

Where \vec a is the acceleration of the object. Note both the force and the acceleration are vectors. The relationship between them is the mass, a scalar.

This means the net force and the acceleration have the same direction.

The sailboat has a mass m= 350 Kg and moves at a=0.62 m/s^2. The magnitude of the net force acting on the boat is:

F=0.62 * 350=217\ N

As stated above its direction is the same as the acceleration, thus:

The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.

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What would happen if an automobile moving at a high speed suddenly comes to a stop? How does this relate to Newton's laws and wh
insens350 [35]
If an automobile moving at high speed suddenly comes to a stop, you would have a large change in momentum. This relates to Newton's second law in the form F = delta p / delta t, where p is momentum (mv).
You could lessen the effect of the sudden stop on the passengers by changing the average force exerted on them. If you look at Newton's second law again, you can see that given some delta p, you can decrease F by increasing delta t. What this means is that if you increase the length of time over which the change in momentum occurs, you can decrease the average force exerted to obtain that change in momentum. This is the reason why landing on a soft cushion is preferable to landing on a concrete surface. The cushion gives way to any object falling on it while still providing some resistance (you don't stop as abruptly), so while your change in momentum is the same in both cases, you have a larger delta t in the case of the cushion.
6 0
3 years ago
what will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.30 meters in the downwar
Anna35 [415]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
Below are the choices the can be found elsewhere:

A.) 14 newtons upward 
<span>B.) 45 newtons upward </span>
<span>C.) 67 newtons upward </span>
<span>D.) 130 newtons upward </span>
<span>E.) 150 newtons upward
</span>
The answer is A.) 14 newtons upward


7 0
3 years ago
Assume that Parker Company will receive SF200,000 in 360 days. Assume the following interest rates:
Anna35 [415]

Answer:

b.  $96,914

Explanation:

360-day borrowing rate = 5%

spot rate = 0.48

360-day deposit rate  = 6%

Borrow at the rate of 5% to get

SF200,000/1.05 = $190,476.19

Convert at the spot rate of $0.48 to get

190,476.19*0.48 = $91,428.57

Invest at the interest rate of 6% to get

91,428.57/1.06 = 96,914.28

Therefore, Parker Company will receive $96,914 in 360 days.

7 0
3 years ago
An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car
saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

F = ma

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

-10000 =2000\times a

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

a = -\dfrac{10000}{2000}

a=-5\ m/s^2

For the distance,

Using third equation of motion

v^2-u^2=2as

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

0-30^2=2\times(-5)\times s

s = 90\ m

Hence, The car traveled the distance before stopping is 90 m.

6 0
3 years ago
What is the frequency of the electromagnetic wave if the period is 1.54x10-15s
pogonyaev

answer✿࿐

I was not able to write it here

so I did it somewhere else and attached the picture

i hope it helps

have a nice day

#Captainpower

3 0
2 years ago
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