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3 years ago

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

5 0

3 years ago

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

5 0

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

4 0

3 years ago

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