All categories

3 years ago

I need your opinion on this

1 answer:

5 0

Block that fool, move on, take a nice warm bath, but on some music and vibe dawg, it might be hard to get over something like this, but just know that person was in the wrong the whole time.

You might be interested in

A toy car with a mass of 2.5 kg travels at a velocity of 37 m/s for 2.5 s until it is stopped by a wall. What was the average ac

kirza4 [7]

I want to say it's A Hope it helps

5 0

3 years ago

Read 2 more answers

The gravitational potential energy of an object is equal to its weight multiplied by its

Elina [12.6K]

Answer:

Height above a surface

Explanation:

Gravitational potential energy is the energy which an object possesses due to its position above a surface.

It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.

It is given mathematically as:

P. E. = m*g*h

where m = mass of the body

g = acceleration due to gravity

h = height above a surface

m*g represents the weight of the object.

Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.

3 0

4 years ago

Date

Sphinxa [80]

Explanation:

Given: time= 3 minutes

Distance =18 km

Speed = distance / time

= 18/3

=6 km/min

(I don't know in what unit u want the answer)

Hope this helps and answers your question..

Good Luck

6 0

3 years ago

A speedboat travels from the dock to the first buoy, a distance of 20 meters, in 18 seconds. It began the trip at a speed of 0 m

LenaWriter [7]

So as a fact we do not know if the boat "burned tires" to reach the buoy :) the average velocity is +1,11 m/s. 20m divided by 18 seconds is +1.11 m/s

5 0

3 years ago

Read 2 more answers

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

3 years ago