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Irina18 [472]
3 years ago
15

How many joules of kinetic energy does a pendulum have when it has 100 joules of potential energy

Physics
2 answers:
Juliette [100K]3 years ago
6 0

Answer :100J of KE.

Explanation:

zloy xaker [14]3 years ago
4 0

Answer:

The maximum kinetic energy is 100 j.    

Explanation:

<h3>The kinetic energy = (potential energy) + (kinetic energy) and the potential energy of 0 J implying its kinetic energy is 100 J, which is its maximum. </h3>

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A 64.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxyg
Umnica [9.8K]

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6 0
3 years ago
1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?
Korvikt [17]

For this case we have that by definition, the momentum equation is given by:

p = m * v

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

m = 0.15 \ kg\\v = 120 \frac {m} {s}

Substituting:

p = 0.15 * 120\\p = 18 \frac {kg * m} {s}

On the other hand, if we clear the variable "mass" we have:

m = \frac {p} {v}

According to the data we have:

p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg

Thus, the mass is 50 \ kg

Answer:

p = 18 \frac {kg * m} {s}\\m = 50 \ kg

3 0
3 years ago
A 22.0 ohm and 75.0 ohm resistor are in parallel, connected to a 5.00 v battery. how much current flows out of the battery
Julli [10]

Answer:

294 mAmps

Explanation:

The two resistors are equal to   22 * 75 / ( 22+ 75) = 17.01 ohm

V = IR

V/R = I

5 / 17.01 = .294 A

5 0
2 years ago
A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma
Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}

in this case:

3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s

the initital kinetic energy is:

K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ

and the final:

K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ

The energy lost is given by:

E_l=|K_f-K_i|\\E_l=23kJ

8 0
3 years ago
Can someone help me with these two please.
Alex17521 [72]
Could you scroll up a little so I can see the first part of the question- I MIGHT be able to help you then. Thanks!
3 0
3 years ago
Read 2 more answers
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