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Lerok [7]
3 years ago
8

What is the equation to find an angle of projectile

Physics
1 answer:
Ksju [112]3 years ago
3 0
I do not recall the answer to this question
You might be interested in
If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?
Elina [12.6K]
Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.
5 0
3 years ago
What is the specific heat of an unknown substance if 2000 J of energy are required to raise the temperature of 4 grams of the su
Dmitriy789 [7]
The energy required to heat a substance is related by the formula:
Q = mCpΔT ; where Q is the energy, m is the mass of the substance, Cp is the specific heat capacity and ΔT is the change in temperature.
2000 = (4)(Cp)(5)
Cp = 100 Joules / g °C
7 0
3 years ago
When you hammer a nail into wood, the nail heats up. 30 Joules of energy was absorbed by a 5-g nail as it was hammered into plac
Goryan [66]

Answer:

13.33 K

Explanation:

Given that,

Heat absorbed, Q = 30 J

Mass of nail, m = 5 g = 0.005 kg

The specific heat capacity of the nail is 450 J/kg-°C.

We need to find the increase in the temperature during the process. The heat absorbed in a process is as follows:

Q=mc\Delta T\\\\\Delta T=\dfrac{Q}{mc}\\\\\Delta T=\dfrac{30}{0.005\times 450}\\\\=13.33\ K

So, the increase in temperature is 13.33 K.

7 0
3 years ago
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.09 mm apart and position
kow [346]

Answer:

pretty dark

Explanation:

i dont know

4 0
3 years ago
PLEASE HURRY
Sophie [7]

Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)

• The net force in the parallel direction is

∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

• The net force in the perpendicular direction is

∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0

Solving the second equation for <em>n</em> gives

<em>n</em> = <em>mg</em> cos(21°)

<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)

<em>n</em> ≈ 1.83 N

Then the magnitude of friction is

<em>f</em> = <em>µn</em>

<em>f</em> = 0.25 (1.83 N)

<em>f</em> ≈ 0.457 N

Solve for the acceleration <em>a</em> :

-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>

<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)

<em>a</em> ≈ -5.80 m/s²

so the block is decelerating with magnitude

<em>a</em> = 5.80 m/s²

down the ramp.

5 0
3 years ago
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