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2 years ago

A man is holding a 5kg weight ball. What is the weight of the ball?

Physics

1 answer:

jonny [76]2 years ago

3 0

Answer:

1000kg

Explanation:

I'm smart I am a genius I know everything so I know it is 1000kg

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A ball rolls from 10 m to -25 m in 2.5 seconds. What was<br> its average velocity?<br> (Units = m/s)

marissa [1.9K]

Answer:

v = -14 m/s

Explanation:

Given that,

Initial location of the ball, X₁ = 10 m

Final position of the ball, X₂ = -25 m

Time taken to travel is, t = 2.5 s

The average velocity of the ball is given by the formula,

V = X₂ - X₁ / t m/s

Substituting the values in the above equation,

V = -25 - 10 / 2.5

= -14 m/s

The negative sign in the velocity indicates that ball rolls in the opposite direction.

Hence, the average velocity of the ball is v = -14 m/s

8 0

4 years ago

Express the measurement 0.00000575 into scientific notation.

g100num [7]

Answer: = 5.75 × 10 -6

Explanation:

= 5.75 × 10-6

(scientific notation)

= 5.75e-6

(scientific e notation)

= 5.75 × 10-6

(engineering notation)

(millionth; prefix micro- (u))

= 0.00000575

(real number)

7 0

3 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

3 years ago

Plz do all of it i will give brainlest and thanks to best answer<br> plz do it right

Gelneren [198K]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

Before the experiment, the total momentum of the system is 2.5 kg m/s to the right and the kinetic energy is 5J. After the exper

finlep [7]

Answer:

Option (b) is correct.

Explanation:

Elastic collision is defined as a collision where the kinetic energy of the system remains same. Both linear momentum and kinetic energy are conserved in case of an elastic collision.

Inelastic collision is defined as a collision where kinetic energy of the system is not conserved whereas the linear momentum is conserved. This loss of kinetic energy may due to the conversion to thermal energy or sound energy or may be due to the deformation of the materials colliding with each other.

As given in the problem, before the collision, total momentum of the system is $2.5~Kg~m~s^{-1}$ and the kinetic energy is $5~J$ . After the collision, the total momentum of the system is $2.5~Kg~m~s^{-1}$ , but the kinetic energy is reduced to $4~J$ . So some amount of kinetic energy is lost during the collision.

Therefor the situation describes an inelastic collision (and it could NOT be elastic).

5 0

3 years ago