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Vera_Pavlovna [14]
3 years ago
6

How heavy is an object that displaces 400N of water in a pool?

Physics
1 answer:
julia-pushkina [17]3 years ago
7 0
All we can say is that the object's volume is about 41 liters. That's the same as the volume of water displaced.

We can't say anything about the object's weight. There is no direct connection between the weight of the object and the weight of the water it displaces.
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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0
3 years ago
Read 2 more answers
Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat
aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R    ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0
2 years ago
If the spring constant is doubled , what value does the period have for a mass on a spring?
zhuklara [117]

Answer:

D. The period would decrease by sqrt (2)

Explanation:

The period of a mass-spring system is given by:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass

k is the spring constant of the spring

If the spring constant is doubled,

k' = 2k

So the new period will be

T'=2\pi \sqrt{\frac{m}{(2k)}}=\frac{1}{\sqrt{2}}(2\pi \sqrt{\frac{m}{k}})=\frac{T}{\sqrt{2}}

So the correct answer is

D. The period would decrease by sqrt (2)

8 0
3 years ago
Read 2 more answers
The radius of the earth is 6378 km. What is the diameter of the earth in meters?
Nesterboy [21]
To determine the diameter of the earth in metres first multiply the original value by 2.

6378 X 2 = 12 756 km.

Then convert km - m

1 km = 1000 m
12 756 km = ? m

12 756 • 1000 = 12 756 000 = 12 756 000 m or 1.2756 X 10 ^ 7 m

The final solution for the diameter is 1.2756 X 10 ^ 7 m.
7 0
3 years ago
URGENTEEEEEEEEEEEEEE
OverLord2011 [107]

Answer:

no habl

Explanation:

6 0
2 years ago
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