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nikdorinn [45]
2 years ago
7

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static fric

tion between the crate and the floor is 0.57. Express your answer using two significant figures.
Physics
1 answer:
fiasKO [112]2 years ago
3 0

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

Where μ = coefficient of friction

m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

Fr = -0.57 * 32 * 9.8

Fr = - 178.75 N

Hence, the force necessary to move the crate must be at least equal to but opposite in direction to this frictional force.

Therefore, this force is 178.75 N

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Identify the origins of breakdown when using a spectrum analyzer
Zanzabum

Four regions of the electromagnetic spectrum that astronomers use when observing objects in the space are the following enumerated answers.

1. First is Ultraviolet

2. Next is Infrared

3. Then the radio

4. Lastly the Visible lights.

These are the answers to the problem.

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2 years ago
What type of properties would a scientist need to know to correctly place in new element into a group?
V125BC [204]
Placing elements into columns, groups, rows and periods that share certain properties. properties such as gas, solid and liquid determine an element physical state at room temperature. 
3 0
3 years ago
A car accelerates from zero to a speed of 110
Verizon [17]

The car's rate of  acceleration : a = 2.04 m/s²

<h3>Further explanation</h3>

Given

speed = 110 km/hr

time = 15 s

Required

The acceleration

Solution

110 km/hr⇒30.56 m/s

Acceleration is the change in velocity over time

a = Δv : Δt

Input the value :

a = 30.56 m/s : 15 s

a = 2.04 m/s²

3 0
3 years ago
A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t
iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

T = 2\pi \sqrt{\frac{l}{g} }

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

T' = 2\pi \sqrt{\frac{2l}{g} }

T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }

T' = \sqrt{2} * T

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0
1 year ago
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
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