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2 years ago

How high above the ground would a 2 kg object need to be in order to have 180 J

Physics

1 answer:

ankoles [38]2 years ago

3 0

Answer:

energy= MGH

2*9.8*h=180

h=180/19.6

h=9.32 m

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Learning Task 2: Write the words that can be associated with the ''Music During Classical Era''. You may ask help from the membe

Natasha_Volkova [10]

Answer:

classical

concert

exposition

orchesta

devolopment

recapitulation

sympathy

sonata

sinfonia

soloist

Explanation:

5 0

3 years ago

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

3 years ago

ball A of mass 5.0 kg moving at 20 meters per second collides with ball B of unknown mass moving at 10. meters per second in the

antiseptic1488 [7]

Answer:The mass of ball B is 10 kg.

Explanation;

Mass of ball A = $M_A=5 kg$

Velocity of the ball A before collision: $U_A=20 m/s$

Velocity of ball A after collision= $V_A=10 m/s$

Mass of ball B= $M_B$

Velocity of the ball B before collision: $U_B=10 m/s$

Velocity of ball B after collision= $V_B=15 m/s$

$M_AV_A+M_BV_B=M_AU_A+M_BU_B$

$5 kg\times 10 m/s+M_B\times 15=5 kg\times 20m/s+M_B\times 10m/s$

$M_B=10kg$

The mass of ball B is 10 kg.

8 0

2 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

3 years ago

If the pendulum took longer to complete one oscillation, how would the graph change?

Sindrei [870]

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

7 0

3 years ago

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