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NNADVOKAT [17]
3 years ago
8

A 35 N force makes a 10 degree angle with the positive x-axis. What is the magnitude of the vertical component of the force?

Physics
2 answers:
Snowcat [4.5K]3 years ago
5 0

Answer:

6.07 N

Explanation:

Given that,

Force, F = 35 N

It makes 10 degree angle with the positive x-axis.

We need to find the magnitude of the vertical component of the force. It can be given by :

F_y=F\sin\theta\\\\=35\times \sin(10)\\\\=6.07\ N

So, the magnitude of the vertical component of the force is 6.07 N.

Mariulka [41]3 years ago
5 0

Answer:

Vertical component of force is equal to 6.067 N

Explanation:

The magnitude of the vertical component on any force when α is the angle between the direction of force and the X -axis is

F_y = F  * Sin\alpha

Substituting the given values in above equation, we get -

F_y = 35  * Sin 10\\F_Y = 35 *  0.1736\\F_Y = 6.067

Vertical component of force is equal to 6.067 N

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A moving car skids to a stop with the wheels locked across a level roadway. Of the forces listed, identify which act on the car.
Vesnalui [34]

Answer:

Normal, Gravity, Friction, and Air Resistance.

Explanation:

When a moving car skid to stop and its wheels are locked across, then the following forces will be applied on the car:

<u>Normal force:</u> It will act counter to gravity that pushes an object against a surface and acts perpendicular to the contact surface.

<u>Gravity:</u> Gravity force acts in each and every object having mass and it can not be avoidable. So, the gravity force will also apply to the car and attract it to the earth's surface.

<u>Friction: </u>Friction is a force that acts opposite to the motion and stops or slows motion. Friction will be applied to the car that will oppose the motion of the car and stop it.

<u>Air resistance:</u> air resistance is defined as the forces exerted by air that acts opposite to the relative motion of an object. Air resistance will also be applied to the car when it will skid to stop as we are always surrounded by the air.

Hence, the correct answers are "Normal, Gravity, Friction, and Air Resistance."

4 0
3 years ago
How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?
lesya [120]
Force = mass * acceleration = 1500kg * 8m/s²
3 0
3 years ago
If it requires 7.0 j of work to stretch a particular spring by 2.1 cm from its equilibrium length, how much more work will be re
SVEN [57.7K]
4.6 j more. To get this take 7 and multiply it by 3.5 to get 24.5 take the x which is what you’re looking for and multiply it by the 2.1 to get 2.1x. Take 24.5 and divide it by 2.1 x and get 11.6. Subtract 11.6 by 7 and get 4.6
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3 years ago
Could anyone can help me for this question I don’t understand how I can do?!! Please it really important!!
stiv31 [10]
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6 0
4 years ago
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

8 0
3 years ago
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