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konstantin123 [22]
3 years ago
7

How much potential energy does a 20-kg barrel have when it is in the truck whose floor is 5.0 m above the ground? P.E. = J

Physics
1 answer:
Morgarella [4.7K]3 years ago
8 0

Answer:

980J

Explanation:

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What is newton's 3rd law of physics ​
ValentinkaMS [17]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

4 0
3 years ago
Due to efficiency considerations related to its bow wake, the supersonic transport aircraft must maintain a cruising speed that
givi [52]

Answer:

decreases.

Explanation:

When the aircraft is flies from the warm air into the  colder air then its speed will be decreases.

as we know that

we know mach number is constant  

so that here Mach number M is expressed as  

M = \frac{u}{v}      .............................1

here u is  Local flow velocity with respect to the boundarie and v is the speed of sound in the medium

If the aircraft flies from hot air to cold air, the speed of sound in the medium will decrease. But the Mach number remains constant. Therefore, the local flow velocity relative to the boundaries also decreases.

7 0
3 years ago
Some children are underfed and often have no place to sleep. According to
jeka94

Answer:

Physiological needs

Explanation:

3 0
3 years ago
What force is needed to move a 5 kg mass with an acceleration of 5 m/s²?
ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

4 0
2 years ago
Read 2 more answers
A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil
mrs_skeptik [129]

Answer:

0.25 L

Explanation:

P_1 = Initial pressure = 1 atm

T_1 = Initial Temperature = 20 °C

V_1 = Initial volume = 4.91 L

P_2 = Final pressure = 5.2 atm

T_2 = Final Temperature = -196 °C

V_2= Final volume

From ideal gas law we have

\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\\Rightarrow P_2=\dfrac{1\times 4.91(273.15-196)}{(20+273.15)\times 5.2}\\\Rightarrow V_2=0.24849\ L\approx 0.25\ L

The pressure experienced by the balloon is 0.25 L

7 0
4 years ago
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