All categories

2 years ago

What happens to the density if the volume of an object decrease and the mass remains the same?

1 answer:

7 0

The density would increase because you still have the same amount of weight, but it is just packed more tightly in a smaller object.

You might be interested in

I need help with two more Physics problems

sesenic [268]

Answer:

c. They hit at the same time

b. BGS

Explanation:

A marble dropped (initial vertical velocity is 0) will land at the same time as a marble launched horizontally (initial vertical velocity is 0) from the same height.

Boat S has a net speed of 5 m/s (10 − 5).

Boat B has a net speed of 15 m/s (10 + 5).

Boat G has a net speed of ≈11.2 m/s (√(10² + 5²)).

8 0

2 years ago

Read 2 more answers

A person swims to the other end of a 20m long pool and back. What is their displacement?

mafiozo [28]

Answer:

Zero

Explanation:

It is given that,

A person swims to the other end of a 20m long pool and back.

We need to find his displacement.

Displacement = shortest path covered

He reaches at the same position as from where he has started. It means the shortest path covered is equal to 0 i.e. his displacement is zero.

3 0

3 years ago

The technological challenges of installing cellphones across the globe

Leona [35]

Answer:

I don't know sorry For this question

3 0

2 years ago

The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration

Alisiya [41]

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

Force = mass x acceleration

Since acceleration of the cart is unknown;

Acceleration = $\frac{force}{mass}$ = $\frac{411}{281}$

Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

8 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago