Suppose you are pushing on a crate across a floor as shown below. Assume the friction force is 47.0 N. How much time will it take for the crate to reach 6.0 m/s if it started from rest? Assume the weight of the crate is 2058 N. (250 N force applied)(Question is no longer a priority but i’d like to know the answer and how it’s found) pls don’t scam i’m serious man i need to know

Answers

Answer 1

Answer:

Answer:

6.2 seconds

Explanation:

Using Newton's second law, ∑F=ma, we know the net force acting on the object is Force applied-Force of friction. The net force is 203 N. Newton's second law requires the mass of an object, not the weight force, so we will have to calculate the mass. We know that m*g=weight force, in this case, solve for the mass and you will get 210 kg. Now that we have the value of the net force and the mass, we can solve for acceleration. $(F)/(m)=a$ =0.967 m/s^2. Now, since we have the acceleration, initial velocity(0 m/s), and the final velocity (6m/s) we will use these to solve for time using the kinematic equation Vf=Vi + at. Plug in the values we know and solve for time and you will get 6.2 seconds

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An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

Answers

Given :

∅ = 60⁰

u = 4 m/s

g = 10m/s²

to find :

T = ?

Solution :

as per formula,

$t = (2u \: sin \theta)/(g)$

now put the value : $t \: = (2 * 4 * sin \: 60)/(10)$

as we know $sin60 \: = ( √(3) )/(2)$

therefore,

$t \: = (8 * ( √(3) )/(2) )/(10)$

as we solve this we get,

$t \: = ( 2√(3) )/(5)$

that's t = 0.69 sec

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0.8 seconds

Explanation:

time of flight = 2u/g

u=4m/s

g=10

= 8/10

= 0.8 sec

just a trial...not sure!!!

The cycle is a process that returns to its beginning, but it does not repeatitself.
True
False

Answers

False. It does repeat itself

The planet uranus is tilted nearly on its side so that its axis or rotation is only 8 degress abway from its orbit plane. if you lived at latitude 45 degrees on uranus for what fraction of the uranian year would answer?

Answers

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

When a box is placed on an inclined surface with no friction, it will:

Answers

Answer: With no friction, the box will accelerate down the ramp

Explanation:

It will gain speed down the ramp

A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.8 m/s, what is the coefficient of static friction? a. 3.0 b. 0.15 c. 0.28 d. 0.31

Answers

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

Based on the calculatins, the coefficient of static friction is equal to: D. 0.31.

Given the following data:

Force = 750 Newton.

Mass = 250 kg.

Acceleration due to gravity = 9.8 $m/s^2$

How to calculate the coefficient of static friction.

Mathematically, the staticfrictional force acting on an object is giving by this formula:

$F_s=\mu N=\mu mg\n\n\mu =(F_s)/(mg)$

Where:

$\mu$ is the coefficient of static friction.

N is the normal force.

m is the mass.

g is the acceleration due to gravity.

Substituting the given parameters into the formula, we have;

$\mu =(750)/(250 * 9.8)\n\n\mu =0.31$

Read more on force here: brainly.com/question/1121817

Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.

Answers

Answer:

2.05 x 10^8 m /s

Explanation:

c = 3 x 10^8 m/s

μ = c / v

where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.

μ = 1.461

1.461 = 3 x 10^8 / v

v = 3 x 10^8 / 1.461

v = 2.05 x 10^8 m /s

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