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garri49 [273]
3 years ago
14

A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle.

Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?
Physics
1 answer:
sertanlavr [38]3 years ago
5 0

Answer:

Explanation:

Normal force of the surface on the box will be

N = mg - Fsinθ

Ν = 10(9.8) - 600sin37

N = -263

As normal force cannot be less than zero, the applied force lifts the crate off the surface.

Now it's just a matter of finding the acceleration

In the horizontal direction, the acceleration is

a = F/m

a = (600cos37) / 10

a =  47.9181... m/s²

the crate weight is mg = 10(9.8) = 98 N.

In the vertical direction the acceleration is

a = ((600sin37 - 98) / 10)

a = 26.3089... m/s²

total acceleration is

a = √(47.9181² + 26.3089²)

a = 54.6653... m/s²

s = ½at²

t = √(2s/a)

t = √(2(1.0)/54.6653)

t = 0.19127...

t = 0.19 s

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The gravitational attraction between a 20 kg cannonball and a 0.002 kg
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2.966\times 10^{-11}\ N

Explanation:

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Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = 6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

F=\dfrac{GMm}{d^2}

Plug in the given values and solve for 'F'. This gives,

F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})

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Read 2 more answers
A roller-coaster car has a mass of 1080 kg when fully loaded with passengers. As the car passes over the top of a circular hill
Dmitriy789 [7]

Answer:

(a): The normal force on the car  from the track when the car's speed is v= 7.6 m/s  is  FN= -6696 N.

(b): The normal force on the car from the track when the car's speed is v= 17 m/s is FN= 8912.7 N.

Explanation:

m= 1080 kg

r= 16m

v1= 7.6 m/s

v2= 17 m/s

g= 9.81 m/s²

v1= w1*r

w1= v1/r

w1= 0.475 rad/s

ac1= w1² * r

ac1= 3.61 m/s²

FN= m * (ac1 - g)

FN= -6696 N    (a)

-----------------------------------------------------

v2= w2*r

w2= v2/r

w2= 1.06 rad/s

ac2= w2² * r

ac2= 18.06 m/s²

FN= m * (ac2 - g)

FN= 8912.7 N    (b)

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