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3 years ago

R: Law of Conservation of Energy Pre-Write

Physics

1 answer:

Bad White [126]3 years ago

3 0

Answer:

V at C is 3.6 m/s

Explanation:

At A kinetic energy is zero and potential energy=mgh=0.5*9.81*0.6=2.943 J

By conservation of energy.

KE+PE=Constant

At C PE=0.6 J

the KE=2.943-0.6=2.343 J

KE=0.5*m*v^2

v=√[KE/(0.5*m)]=3.06 m/s

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10. How does surface area affect the amount of air resistance an object experiences?

ikadub [295]

Answer:cross-sectional area, and thus surface area, increases the amount of air resistance an object experiences

Explanation:

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3 years ago

A camera lens with focal length f = 50 mm and maximum aperture f>2

Brut [27]

Answer:

The minimum distance between two points on the object that are barely resolved is 0.26 mm

The corresponding distance between the image points = 0.0015 m

Explanation:

Given

focal length f = 50 mm and maximum aperture f>2

s = 9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D

Substituting the given values, we get –

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5

Y’ = 0.0015 m

8 0

3 years ago

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor

joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is $\Delta U_1=222 J$

Work done on the system $W_1=-150 J$

For second step

change in internal Energy of the system is $\Delta U_2=123 J$

Work done on the system $W_2=-195 J$

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

$Q=\Delta U+W$

for first step

$Q_1=222-150=72 J$

$Q_2=123-195=-72 J$

overall heat added $=Q_1+Q_2$

$Q_{net}=72-72 =0$

For overall Process Heat added is 0 J

8 0

3 years ago

What is the displacement of the armadillo between 0s and 24s ?

Ann [662]

Answer:

Displacement: 6 meters

Distance: 24 meters

Explanation:

4 0

3 years ago

A 170 kg astronaut (including space suit) acquires a speed of 2.25 m/s by pushing off with his legs from a 2600 kg space capsule

saw5 [17]

Explanation:

Mass of the astronaut, m₁ = 170 kg

Speed of astronaut, v₁ = 2.25 m/s

mass of space capsule, m₂ = 2600 kg

Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :

initial momentum = final momentum

Since, initial momentum is zero. So,

$m_1v_1+m_2v_2=0$

$170\ kg\times 2.25\ m/s+2600\ kg\times v_2=0$

$v_2=-0.17\ m/s$

So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.

8 0

3 years ago