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bulgar [2K]
2 years ago
5

Calculate the speed of an object that moves 200 m in 4 s?

Physics
1 answer:
Rasek [7]2 years ago
7 0
50m/s

I divided the 200m by 4s (so 200/4) to get 50m per second
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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th
Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, m=\dfrac{F}{a}

m=\dfrac{70\ N}{7.5\ m/s^2}

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

W=m\times g

W=9.34\ kg\times 1.62\ m/s^2

W = 15.13 N

or

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0
3 years ago
Does the sun fuse hydrogen into helium or vice versa?
NikAS [45]

Hydrogen has the smaller, lighter, simpler nucleus.
Helium has the larger, heavier, more complex nucleus.
Hydrogen nuclei are fused to form helium nuclei. 
When that happens, energy is released. 

6 0
2 years ago
The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
If i walk to school everyday will i loose weight?
Sphinxa [80]
You would also have to eat right lol
7 0
3 years ago
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Tim's car breaks down just down the street from his house. he and his friends decide to try to push it to his house. using newto
daser333 [38]
Pushing a broke down car, even done by more than one person, is difficult especially if the distance to be covered is quite far. A car is heavy and it requires a lot of force to start the car moving. This is because the inertia of the car to remain at rest is great. Additionally, the force applied in pushing the car must be greater than the frictional force to cause it to accelerate. The frictional force is dependent on the mass of the object which means that the frictional force acting on the car is also great. Finally, with every push of the car, the frictional force will always be present and acting on the opposite direction. The push that will be supplied must be sustained all throughout.
6 0
3 years ago
Read 2 more answers
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