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2 years ago

What is the net force? URGENT! thank you

Physics

1 answer:

alexandr402 [8]2 years ago

6 0

Net force is the vector sum of forces acting on a particle or body. The net force is a single force that replaces the effect of the original forces on the particle's motion. It gives the particle the same acceleration as all those actual forces together as described by the Newton's second law of motion.

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A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

2 years ago

Consider a large truck and a small car driving up a straight steep hill. The truck is moving at 60 miles per hour and the car at

Oksana_A [137]

Answer:

The correct answer is option A.

Explanation:

Force is defined as push or pull on an object with mass due to which change velocity occurs (acceleration).

$Force=Mass\times Acceleration$

$F=ma$

$a=\frac{dv}{dt}[tex][tex]m\times \frac{dv}{dt}[tex]We are given with large truck and small car at constant speed, which means that acceleration zero.[tex]a=\frac{dv}{dt}=0$ if ,v = constant

Net force on the vehilce

F = Normal - Weight

0 = N = W

N = W

So, the force experienced by the object will be due to its mass, and higher the mass more will be the experienced by an object.So, large truck will experience larger net force.

7 0

2 years ago

Light is incident on a piece of glass in air at an angle of 33 degrees from the normal. If the index of refraction of the glass

lilavasa [31]

About 21 to 22 degrees

as below

4 0

2 years ago

An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?

Kamila [148]

Answer:

The answer is

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

$m = \frac{p}{v} \\$

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

$m = \frac{280}{10} = 28 \\$

We have the final answer as

Hope this helps you

3 0

2 years ago

A machine

In-s [12.5K]

Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

$W = F*d$

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

$W = (80*10)*3\\W = 2400 [J]$

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

$P=\frac{W}{t}$

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

$P = 2400/40\\P = 60 [W]$

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

$Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]$

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2 years ago