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sweet-ann [11.9K]
3 years ago
6

If a 5.00 kg box slides down a ramp inclined at 60.0° above the horizontal, what is the

Physics
1 answer:
Basile [38]3 years ago
8 0

Answer:

A 70 kg box is slid along the floor by a 400 n force. The coefficient of friction between the box and the floor is 0. 50 when the box is sliding

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Jane is sliding down a slide. What kind of motion is she demonstrating? A. translational motion B. rotational motion C. vibratio
Andreyy89
A. translational motion
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One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large
Eva8 [605]

Answer:

6666.67 Newtons

Explanation:

The formula F=ma (force is equal to mass multiplied by acceleration) can be used to calculate the answer to this question.

In this case:

  • mass= 0.1mg= 1*10^-7 kg
  • velocity= 4.00*10^3 m/s
  • time= 6.00*10^-8 s

Using velocity and time, acceleration can be calculated as:

  • a= 6.667*10^10 m/s²

Substituting these values into the formula F=ma, the answer is:

  • F= (1*10^-7)kg * (6.667*10^10) m/s²
  • F= 6666.67 Newtons of force
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3 years ago
Which option lists a form of kinetic energy followed by a form of potential
Artemon [7]

Answer:

D. Sound Energy, Magnetic energy

Explanation:

Sound energy is in motion, and Magnetic energy is about to be in motion.

6 0
3 years ago
Why does a round pizza come in a square box? Random questions
denpristay [2]

Answer:

because they dont know how big it willactually be? idk

or so deep dish can fit too

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Read 2 more answers
1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g
Delvig [45]

1) 6.67\cdot 10^{-4} N

The force of gravitation between the two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where

G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2} is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N


2)  2.7\cdot 10^{-3} N

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

F \sim \frac{1}{r^2}

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F

So, the gravitational force increases by a factor 4. Therefore, the new force will be

F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N


3)  12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

\tau=Fd

Where, in this case:

F = 25 N is the perpendicular force

d = 0.5 m is the distance between the force and the center

By using the equation, we find

\tau=(25 N)(0.5 m)=12.5 Nm


4) 0.049 kg m^2/s

The relationship between angular momentum (L), moment of inertia (I) and angular velocity (\omega) is:

L=I\omega

In this problem, we have

I=0.007875 kgm^2

\omega=6.28 rad/s

So, the angular momentum is

L=I\omega=(0.007875 kgm^2)(6.28 rad/s)=0.049 kg m^2/s

6 0
3 years ago
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