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WITCHER [35]
3 years ago
5

Different objects don't heat at the same rate because they have different

Physics
1 answer:
Kaylis [27]3 years ago
7 0
Specific heat

Let me know if you need further explanation!
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Three pendulums all have the same length and start from the same height. The first pendulum is very light and has a mass of 67 g
vovikov84 [41]

Answer:

All three pendulum will attain same velocity

Explanation:

All three pendulum will attain same velocity irrespective of their mass difference in isolated system (means where air drag are negligible) and at same length

As you know when velocity is calculated we can not take mass into account.  

3 0
3 years ago
A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.
Kay [80]

Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g=0.27\times 10^{-3} kg

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm=8.5\times 10^{-2} m

1m=100 cm

a.Potential energy stored in the compressed spring  is given by

P.E=\frac{1}{2}kx^2

P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2

P.E=6.5025 J

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

6 0
3 years ago
What is the pressure 100m below the surface of the sea , if the density of sea water is 1150kg/m3
Olin [163]
The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
4 0
3 years ago
A small plane starts from rest and accelerates uniformly to the east to a takeoff velocity of 70 m/s in 5 seconds. What is the p
Rainbow [258]
Hope this helps, if you need clarification i got you

3 0
3 years ago
A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at
pav-90 [236]

Answer:

Explanation:

Given

mass of drop m=2 kg

height of fall h=1 m

ball leaves the foot with a speed of 18 m/s at an angle of 55^{\circ}

Velocity of ball just before the collision with the floor

u^=2gh

u=\sqrt{2gh}

u=\sqrt{2\times 9.8\times 1}=4.42 m/s

Impulse delivered in Y direction

J_y=m(v\sin (55)-(-u))

J_y=2(18\sin (55)+4.42)

J_y=38.32 kg-m/s

Impulse in x direction

J_x=m\times v\cos (55)

J_x=2\times 4.42\cos (55)=20.646

J_{net}=\sqrt{J_x^2+J_y^2}

J_{net}=\sqrt{(38.32)^2+(20.64)^2}

J_{net}=43.52 kg-m/s

at an angle of \tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}

\phi =tan^{-1}(1.856)

\phi =61.7^{\circ}  

7 0
3 years ago
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