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oksano4ka [1.4K]
2 years ago
7

The coefficients of static and kinetic frictions for plastic on wood are 0.50 and 0.40,respectively. How much horizontal force w

ould you need to apply to a 3.0 N plastic calculatorto start it moving from rest?
Physics
1 answer:
IrinaK [193]2 years ago
5 0

The horizontal force needed to start the calculator moving from rest is 1.5 N

What is Kinetic friction?

It is defined as a force that acts between moving surfaces.

The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.

Here,

weight of calculator, N = 3 N

The coefficients of static frictions, µ (static) = 0.50

The coefficients of kinetic frictions, µ (kinetic) = 0.40

Now,

The horizontal force required = The static friction force

F = µ (static) * weight of calculator

F = 0.50 * 3.0

F = 1.5 N

Hence,

The horizontal force needed to start the calculator moving from rest is 1.5 N

Learn more about   horizontal force here:

<u>brainly.com/question/21481680</u>

<u />

#SPJ4

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In 2007, michael carter (u.s.) set a world record in the shot put with a throw of 24.77 m. what was the initial speed of the sho
Serga [27]

The initial speed of the shot is 15.02 m/s.

The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.

Pl refer to the attached diagram.

Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.

Write an expression for R.

R=u_xt=(ucos \theta)t

Therefore,

t=\frac{R}{ucos\theta} .......(1)

In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




7 0
3 years ago
A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a
Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

\Delta x=v_0t+\frac{gt^2}{2}

In this case, the sphere starts from rest, so v_0=0. Replacing the given values and solving for g':

g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}

The acceleration due to gravity near Earth's surface is g=9.8\frac{m}{s^2}. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0
3 years ago
Approximately what percentage of incoming solar radiation is absorbed by the oceans and continents
AleksandrR [38]
90 percent a day to keep things running smoothly
8 0
3 years ago
A solid weighs 16.5N on the surface of the moon. The force of gravity on the moon is 1.7N/Kg.
Fiesta28 [93]

Answer:

mass = 9.7 kg

Explanation:

Weight = Mass x Acceleration due to gravity (g)

16.5 = mass x 1.7

mass = \frac{16.5}{1.7} = 9.7 kg

3 0
2 years ago
What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2
jasenka [17]

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

4 0
2 years ago
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