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olya-2409 [2.1K]
3 years ago
5

Describe the conversions between potential and kinetic energy that occur when you shoot a basketball at a basket

Physics
1 answer:
Aleks04 [339]3 years ago
6 0
When you shoot a basketball at a basket, the energy from your hands is transferred into the basketball. When you hold the ball up in preparation to shoot, it has potential energy, meaning that it has stored energy that can turn into kinetic energy. When you shoot it, that potential energy turns into kinetic energy, causing the basketball to move.
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A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t
user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) =  m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg *  0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0
4 years ago
75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0
3 years ago
Particle A has very little mass in comparison to Particle B. Both particles are in the same atom. Which is the best conclusion a
lesya692 [45]
I think the correct answer would be <span>Particle A orbits the nucleus, and Particle B is located in the nucleus. I think Particle A represents the electrons while B represents protons and neutrons. Hope this answers the question. Have a nice day.</span>
5 0
3 years ago
Read 2 more answers
A warm-up is the exact same thing as a flexibility workout<br><br> True<br> False
kykrilka [37]
False :) ✨ got you my g
3 0
4 years ago
Estimate the speed of the water free surface and the time required to fill with water a cone-shaped container 1.5 m high and 1.5
Zolol [24]

Answer:

speed of water is 0.0007138m/s

Explanation:

From the law of conservation of mass

Rate of mass accumulation inside vessel = mass flow in - mass flow out

so, dm/dt = mass flow in - mass flow out

taking p as density

d \frac{dQ}{dt} = pq_i_n

where,

q(in) is the volume flow rate coming in

Q = is the volume of liquid inside tank at any time

But,

dQ = Adh

where ,

A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

r is the radius of liquid surface

r = (1.5/2) \div 1.5 h = \frac{h}{2}

Hence,

\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n

\frac{dh}{dt} = \frac{q_i_n}{\pi (\frac{h}{2})^2 } =\frac{4q_i_n}{\pi h^2}

so, the speed of water surface at height h

v = \frac{dh}{dt} =\frac{4q_i_n}{\pi h^2}

where,

q_i_n is 75.7 L/min = 0.0757m³/min

h = 1.5m

so,

v = \frac{4 \times 0.0757}{\pi \times 1.5^2} \\\\v = 0.04283m/min

v = 0.04283 /60

v = 0.0007138m/s

Hence, speed of water is 0.0007138m/s

5 0
3 years ago
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