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harina [27]
3 years ago
7

A shotputter throws the shot with an initial speed of 15.6 m/s at a 30.0 degree angle to the horizontal. Calculate the horizonta

l distance traveled by the shot if it leaves the athlete's hand at a height of 2.20m above the ground.
Physics
1 answer:
vfiekz [6]3 years ago
8 0
To calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 above the ground we can get the root of the quad equation for time are t=-0.24 or t =1.84 taking the t = 1.84, so the equation will be:
x = 15.6cos(30) * 1.86, x = 24.79m
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4 0
3 years ago
Please I need help........
LenKa [72]

Answer:

7.46 J/kg/K

Explanation:

The heat absorbed or lost is:

q = mCΔT

where m is the mass, C is the heat capacity, and ΔT is the change in temperature.

Given q = 15.0 J, m = 0.201 kg, and ΔT = 10.0 °C:

15.0 J = (0.201 kg) C (10.0 °C)

C = 7.46 J/kg/°C

Which is the same as 7.46 J/kg/K.

7 0
3 years ago
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What happens when an immovable object meets an unstoppable force?”
Alisiya [41]

Answer:

Sorry to tell you this but there’s no answer for that

Explanation:

4 0
3 years ago
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I need help ASAP....
Svetradugi [14.3K]

Answer:

Meters

Explanation:"How FAR did the athlete run?"

Also it talked about meters

5 0
4 years ago
What is the radiation pressure 1.5 m away from a 700 W lightbulb? Assume that the surface on which the pressure is exerted faces
baherus [9]

Answer:

3.30 x 10^-7 Pascal

Explanation:

distance r = 1.5 m

power P = 700 W

the radiation pressure is given as

Pr = P/A*c

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area of the surface A = 4πr^2

calculate for A

speed of light is c = 3×10^8  m/s

plugging above values in equation above gives

Pr = 3.30 x 10^-7 Pascal

3 0
3 years ago
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