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Vlad [161]
2 years ago
15

Pahelp po ako.

Physics
1 answer:
NeTakaya2 years ago
4 0
D bc thats sound like the only resonable answer
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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela
NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

  • x-component of the tension in the rope on the right is approximately <u>91.93 N</u>
  • y-component of the tension in the rope on the right is approximately <u>71.135 N</u>
  • x-component of the tension in the rope on the left is -80.0 N
  • y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

\displaystyle a_x = \frac{F_x}{m}  \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}

  • The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0
2 years ago
What planet has orbital plane different than the rest of planets?
bixtya [17]
The sun has orbited along time so when they ask theses questions I give you the right answer I think lol
4 0
3 years ago
1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force
Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

4 0
3 years ago
What principle explains why lifting heavy objects is easier using ramps?
Anna11 [10]
By using ramps you can easily push or pull the object up the ramp.
5 0
3 years ago
Read 2 more answers
un litro de un gas es calentado a presión constante desde 20°C hasta 60°C que volumen final ocupará dicho gas? ​
WINSTONCH [101]

Answer:

Final volume, V2 = 3 Litres

Explanation:

Given the following data;

Initial volume, V1 = 1 litre

Initial temperature, T1 = 20°C

Final temperature, T2 = 60°C

To find the final volume, we would use Charles' law;

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

V1/T1 = V2/T2

Making V2 as the subject formula, we have;

V1T2 = V2T1

V2 = (V1T2)/T1

Substituting into the formula, we have;

V2 = (1 * 60)/20

V2 = 60/20

Final volume, V2 = 3 Litres

4 0
3 years ago
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