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Alekssandra [29.7K]
2 years ago
6

Three containers are filled with water to the same level. Identical marbles are dropped into all the three containers. In which

container will the water level be the highest?
a. P

b. Q

c. R

d. The level will be the same in all the three containers.​

Physics
2 answers:
dimaraw [331]2 years ago
5 0

Answer:

P

Explanation:

,............................................ I have. a doubt too on D

slamgirl [31]2 years ago
5 0
Answer: D
The answer is d
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A boy lifted a 50 newton rock 1 meter. How much work was done?
dem82 [27]

Answer:

The answer is 50 Nm

Explanation:

<h3><u>Given</u>;</h3>
  • Applied Force = 50 Newton
  • Total Displacement = 1 meter
<h3><u>To </u><u>Find</u>;</h3>
  • Work done = ?

Here,

W = F • d

W = 50 • 1

W = 50 Nm

Thus, Work done is 50 Nm

<u>-TheUnknownScientist 72</u>

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2 years ago
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How much power will be required to force a current of 4.13 amps to flow through a conductor whose resistance is 113 ohms? Use tw
-Dominant- [34]

The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = square current (I²)× resistancet (R)

P = I²R

<h3>How to determine the power</h3>
  • Current (I) = 4.13 A
  • Resistance (R) = 113 ohms
  • Power (P) =?

P = I²R

P = 4.13² × 113

P = 1927.43 watts

Thus, the power required is 1927.43 watts

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2 years ago
the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

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3 years ago
A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh
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26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

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3 years ago
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A stone is thrown vertically upward with a speed of 15.5 m/s from the edge of a cliff 75.0 m high .
rjkz [21]

a) 2.64 s

We can solve this part of the problem by using the following SUVAT equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the stone

u is the initial velocity

t is the time

a is the acceleration

We must be careful to the signs of s, u and a. Taking upward as positive direction, we have:

- s (displacement) negative, since it is downward: so s = -75.0 m

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)

Substituting into the equation,

-75.0 = 15.5 t -4.9t^2\\4.9t^2-15.5t-75.0 = 0

Solving the equation, we have two solutions: t = -5.80 s and t = 2.84 s. Since the negative solution has no physical meaning, the stone reaches the bottom of the cliff 2.64 s later.

b) 10.4 m/s

The speed of the stone when it reaches the bottom of the cliff can be calculated by using the equation:

v=u+at

where again, we must be careful to the signs of the various quantities:

- u (initial velocity) positive, since it is upward: +15.5 m/s

- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2

Substituting t = 2.64 s, we find the final velocity of the stone:

v = 15.5 +(-9.8)(2.64)=-10.4 m/s

where the negative sign means that the velocity is downward: so the speed is 10.4 m/s.

c) 4.11 s

In this case, we can use again the equation:

s=ut+\frac{1}{2}at^2

where

s is the displacement of the package

u is the initial velocity

t is the time

a is the acceleration

We have:

s = -105 m (vertical displacement of the package, downward so negative)

u = +5.40 m/s (initial velocity of the package, which is the same as the helicopter, upward so positive)

a = g = -9.8 m/s^2

Substituting into the equation,

-105 = 5.40 t -4.9t^2\\4.9t^2 -5.40 t-105=0

Which gives two solutions: t = -5.21 s and t = 4.11 s. Again, we discard the first solution since it is negative, so the package reaches the ground after

t = 4.11 seconds.

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3 years ago
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