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3 years ago

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A 12 kg box is at rest on your kitchen counter, which your cat is pawing at with a horizontal force of 40 N. If the coefficient

of static friction is 0.3, then your cat will succeed in knocking the box off the counter. True or false?

1 answer:

murzikaleks [220]3 years ago

8 0

Answer; I think it's False.

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beks73 [17]

Answer:

raise the board to a higher angle

Explanation:

Static friction is the force opposite to the applied force.

Static friction is dependent on the angle of inclination, it means as the angle of incline increases, the force of friction will increases as normal force will decrease.

So, if the board will be raised to a higher angle, it will increase the angle of incline and will overcome the static friction and block will be able slide.

Hence, the correct option is "raise the board to a higher angle".

5 0

3 years ago

A 1 kg object can be accelerated at 10 m/s^2. If you apply this same force to a 4kg object, what will its acceleration be?

AleksAgata [21]

Answer:

$a2 = 2.5~m/s^2$

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force F exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

Assume we apply some given force F to an object of m1=1 Kg that produces an acceleration $a1=10 m/s^2$ , then:

F = m1.a1

The same force F is now applied to a second object m2=4 Kg that produces an acceleration a2, then:

F = m2.a2

Dividing both equations:

$\displaystyle 1=\frac{m1.a1}{m2.a2}$

Solving for a2:

$\displaystyle a2=\frac{m1.a1}{m2}$

Substituting values:

$\displaystyle a2=\frac{1*10}{4}$

$a2 = 2.5~m/s^2$

7 0

3 years ago

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0

3 years ago

What is the difference between rutherford's model of the atom and bohr's model of the atom

Amanda [17]

Answer:

Rutherford described the atom as consisting of a tiny positive mass surrounded by a cloud of negative electrons. Bohr thought that electrons orbited the nucleus in quantised orbits. Bohr built upon Rutherford's model of the atom. ... So it was not possible for electrons to occupy just any energy level.

Explanation:

7 0

3 years ago

List 3 quantities of waves.

VARVARA [1.3K]

If you mean like electromagnetic waves then, Mico waves, UV rays, and infrared waves

8 0

3 years ago

Read 2 more answers