An object with a mass of 5kg accelerates at 2m/s2. How much force in Newtons(N) is needed to cause this to happen?

Object A has a mass of 100 grams. Object B has a mass of 150 grams. They are both traveling at the same velocity. Which can you

Tanzania [10]

Object B has a greater inertia i think, it has a greater momentum anyway

6 0

3 years ago

We’ve already investigated this problem with one spring scale in Think About 34.1. Now, imagine you have two spring scales, A an

Debora [2.8K]

Answer:

Explained

Explanation:

1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2

2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.

5 0

3 years ago

A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart

Musya8 [376]

1) $0.61 m/s^2$

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

$g=\frac{GM}{r^2}$ (1)

where

G is the gravitational constant

$M=5.98\cdot 10^{24} kg$ is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

$r=4R$

where

$R=6.37\cdot 10^6 m$ is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

$g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2$

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

$g=0.61 m/s^2$ is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

$T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s$

4 0

3 years ago

What would happen if an Asteroid hit earth?

svp [43]

Answer:

There would be tremendous earthquakes and tsunamis, followed by massive volcanism around the impact zone. The ozone layer would be destroyed. The oceans would turn acidic. The Sun would be blotted out, probably for decades.

Explanation:

Hope this helps!

4 0

3 years ago

Read 2 more answers

Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s

Gekata [30.6K]

Answer:

The number of crates is 84580.

Explanation:

mass, m = 30 kg

height, h = 0.9 mm

Power, P = 0.5 hp = 0.5 x 746 W = 373 W

time, t = 1 minute = 60 s

Let the number of crates is n.

Power is given by the rate of doing work.

$P = \frac{n m gh}{t}\\\373 =\frac{n\times 30\times9.8\times 0.9\times 10^{-3}}{60}\\\\n =84580$

7 0

3 years ago