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katen-ka-za [31]
2 years ago
11

A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g

=10m/s2, the body will reach the surface of the earth in how much time?

Physics
1 answer:
kap26 [50]2 years ago
7 0
I know you are Indian by your question, HC Verma class 9 or 11 !!

if you got any problem, comment !!

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In ocean waves, water particles move ________ and energy moves ________.
morpeh [17]
In ocean waves, water particles move with mechanical energy and energy moves with gravity

Not sure but hope it helps!
6 0
1 year ago
Read 2 more answers
A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of
qaws [65]

Answer:

\frac{E_1}{E_2}= 4

Explanation:

Capacitance C is given by

C= \frac{\epsilon_0A}{d}

A= area of capacitor cross section

d= distance

therefore,

C_1= \frac{\epsilon_0A_1}{d_1}

A_1= πR^2

d_1= d

C_2= \frac{\epsilon_0A_2}{d_2}

A_= π(2R)^2

d_2 = 2d

q= \frac{\epsilon_0A_1}{d_1}V_1

threfore

V_1= \frac{qd_1}{\epsilon_0A_1}

and

V_2= \frac{qd_2}{\epsilon_0A_2}

also we know that E= V/d

⇒\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}

⇒\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}

= A_1/A_2= \frac{4R^2}{R^2}=4

therefore,

\frac{E_1}{E_2}= 4

5 0
1 year ago
A computer that is 87% efficient consumes 375 kWh of energy. How much useful energy does it provide?
tekilochka [14]

Answer:

326.25 kWh

Explanation:

Efficiency of a machine is defined as the ratio of useful energy to that of the energy consumed by the machine.

Here, efficiency is given as 87% and the energy consumed by the computer is 375 kWh.

Efficiency, \eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}

Plug in the values of \eta=0.87 and 375 kWh for energy consumed. Solve for useful energy. This gives,

Efficiency, \eta=\frac{\textrm{Useful energy}}{\textrm{Energy consumed}}\\ 0.87=\frac{\textrm{Useful energy}}{375}\\ \textrm{Useful energy}=0.87\times 375=326.25 \textrm{ kWh}

Therefore, the useful energy provided by the computer is 326.25 kWh.

3 0
1 year ago
Which radioactive isotope would take the least amount of time to become stable? rubidium-91 iodine-131 cesium-135 uranium-238
Leno4ka [110]
The radioactive isotope that would take the least amount of time to become stable is rubidium-91. This is because this isotope is the most stable compared to the rest. This was determined by subtracting its atomic mass by its atomic number. The isotope with the least number of difference is the most stable, while the one with the greatest difference is the most unstable.

Difference:
Rubidium: 54 (most stable)
Iodine: 78
Cesium: 80
Uranium: 146 (least stable)
8 0
1 year ago
Read 2 more answers
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