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maks197457 [2]
3 years ago
9

Which is the best example of the law of conservation of energy?

Physics
1 answer:
Sonja [21]3 years ago
6 0

Answer:

Water can produce electricity. Water falls from the sky, converting potential energy to kinetic energy.

Explanation:

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Find the equivalent resistance of this parallel circuit with two strands.
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In a parallel circuit, the total resistance calculated from the individual resistances is computed from the formula: 1/Rt = 1/R1 + 1/R2. substituting R1 and R2, then 
1/Rt = 1/7 + 1/49 
1/Rt = 1/6.125 = 1/ 49/8
Rt = 49/8 <span>Ω

The total resistance hence is </span>49/8 Ω
6 0
3 years ago
Which of the following is not an example of a polymer?
Gala2k [10]
Concrete is not a polymer which Nylon, and Kevlar are
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4 years ago
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Much of the energy of falling water in a waterfall is converted into heat. If all the mechanical energy is converted into heat t
Assoli18 [71]

Answer:

Explanation:

potential energy of water is converted into heat energy

P E = mgh = m x 9.8 x 111

= 1087.8 m .

Let rise in temperature θ

m x s x θ = heat absorbed

m x s x θ =  1087.8 m

m x 4182 x θ =  1087.8 m

θ = 0.26 degree celsius .

6 0
3 years ago
A very hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. The ball rebounds back
saw5 [17]

Answer:

The force exerted by the floor is 80 N.

Explanation:

Given that,

Mass of ball = 0.5 kg

Velocity= 4 m/s

Time t = 0.05 s

When the ball rebounds then the kinetic energy is

K.E =\dfrac{1}{2}mv^2

Where, m = mass of ball

v = velocity of ball

Put the value into the formula

K.E=\dfrac{1}{2}\times0.5\times(4)^2

K.E = 4\ J

The average force exerted by the floor on the ball = change in kinetic energy over collision time

F = \dfrac{4}{0.05}

F=80\ N

Hence, The force exerted by the floor is 80 N.

4 0
3 years ago
A 4.00kg counterweight is attached to a light cord, which is would around a spool. The spool is a uniform solid cylinder of radi
Sergio [31]

Answer:

Explanation:

Given that,

Mass of counterweight m= 4kg

Radius of spool cylinder

R = 8cm = 0.08m

Mass of spool

M = 2kg

The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:

Then we have,

τ(net) = R~ × T~

τ(net) = R~•i × mg•j

τ(net) = Rmg• k

τ(net) = 0.08 ×4 × 9.81

τ(net) = 3.139 Nm •k

The magnitude of the net torque is 3.139Nm

b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:

L~ = R~ × m~v + I~ω

L = mRv + MR v

L = (m + M)Rv

L = (4 + 2) × 0.08

L = 0.48 Kg.m

C. τ =dL/dt

mgR = (M + m)R dv/ dt

mgR = (M + m)R • a

a =mg/(m + M)

a =(4 × 9.81)/(4+2)

a = 6.54 m/s

6 0
4 years ago
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