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3 years ago

What force is needed to accelerate an object 5 m/s^2 if the object has a mass of 10kg?

1 answer:

4 0

It is important to pay attention to the fundamental units in the given information of physics problems; if you carefully arrange a system so the units cancel out appropriately, the answer is usually correct.

Force is mass x acceleration, and the units of force are (Kg/m·s²). In this system, a 10 Kg system is accelerated by 5 m/s². Simply multiply:
10 Kg x 5 m/s² = 50 Kg·m/s² = 50 Newtons

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3 years ago

A bird flying straight upward at 5 m/s drops a berry when it is 300 m above the ground. How fast is the berry going when it hits

Nikitich [7]

Answer:

v=77.62 m/s

Explanation:

Given that

h= - 300 m

speed of the bird ,u= 5 m/s

Lets take Speed of the berry when it hit the ground = v m/s

we know that ,if object is moving upward

v² = u² - 2 g h

u=Initial speed

v=Final speed

h=Height

Now by putting the values

v² = u² - 2 g h

v² = 5² - 2 x 10 x (-300) ( take g = 10 m/s²)

v² =25 + 20 x 300

v² ==25 + 6000

v² =6025

v=77.62 m/s

Therefore the final speed of the berry will be 77.62 m/s.

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3 years ago

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

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3 years ago

Read 2 more answers