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3 years ago

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A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates at 1.5m/s2? A. 51.5 N B. 4

8.5 N C. 75 N D. 33.3 N

1 answer:

const2013 [10]3 years ago

3 0

Use the Second Law of Newton, which states that Net Force = mass times acceleration

F = m * a

F = 50 kg * 1.5 m/s^2 = 75 N

Answer: option C

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A mover loads a 100 kg box into the back of a moving truck by

NeX [460]

Answer:

2.7

Explanation:

The following data were obtained from the question:

Mass (m) of box = 100 Kg

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

Mechanical advantage of a ramp is simply defined as the ratio of the length of the ramp to the height of the ramp. Mathematically, it is given by:

Mechanical Advantage = Lenght / height

MA= L/H

With the above formula, we can obtain the mechanical advantage of the ramp as follow:

Length (L) of ramp = 4 m

Height (H) of ramp = 1.5 m

Mechanical advantage (MA) of ramp =?

MA = 4/1.5

MA = 2.7

Therefore, the mechanical advantage of the ramp is 2.7

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3 years ago

How do I do number 8, plz respond quick, I have a big unit test on this and I’m rocking a 65 science average rn. Thank you for t

xxTIMURxx [149]

Answer:

3 is the correct answer for number 8

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3 years ago

Read 2 more answers

If measurements of gas are 75 L and 300 kIlopascals and then the gas is measured is second time and found to be 50 L, describe w

Ann [662]

Answer:

The pressure must have increased in the process

Explanation:

The State Equation for gasses reads: $P*V=n*R*T$

where P is the gas' pressure, V its volume, n the number of moles of gas, R the gas constant and T the temperature in degrees Kelvin.

If the temperature of the gas doesn't change in the described process, the right hand side of the equation stays the same. If that is the case, given that when the Volume of the gas diminishes from 75 liters to 50 liters, then the pressure must have increased to keep that product "P * V" constant:

$P_i*V_i=P_f*V_f\\75 *300=50*X\\X=\frac{75*300}{50} =450$

So the pressure must have gone up to 450 kilopascals.

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2 years ago

What happens to the force between two charges when each charge is doubled and the distance between them is 1/4 its original

DIA [1.3K]

Answer:

F' = 64 F

Explanation:

The electric force between charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

q₁ and q₂ are charges

r is the distance between charges

When each charge is doubled and the distance between them is 1/4 its original magnitude such that,

q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)

New force,

$F'=\dfrac{kq_1'q_2'}{r'^2}$

Apply new values,

$F'=\dfrac{k\times 2q_1\times 2q_2}{(\dfrac{r}{4})^2}\\\\=\dfrac{k\times 4q_1q_2}{\dfrac{r^2}{16}}\\\\=64\times \dfrac{kq_1q_2}{r^2}\\\\=64F$

So, the new force becomes 64 times the initial force.

7 0

2 years ago

A 87 arrow is fired from a bow whose string exerts an average force of 105 on the arrow over a distance of 75 .

timofeeve [1]

The solution would be like this for this specific problem:

V^2 = 2AS = 2FS/M

V = sqrt(2FS/M) = sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps

So the speed of the arrow as it leaves the bow is 42.5 mps.

I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

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3 years ago