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MAXImum [283]
3 years ago
10

A cylindrical tank has a radius of 53 cm and a height of 1200 mm. It weighs 9.6 kN

Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0

Answer:

1985kg

Explanation:

assuming that

pi =3.14

oil density = 950kg/ cubic meter

g= 9.8m/s

M=\frac{9600}{9.8 } + 950 * 3.14*.53^{2} *1.2=1985kg

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When the electrons reach the collector, they flow towards the positivly charged grid. The resulting current is measured. Note th
Masja [62]

Answer:

We can conclude by saying that in the beginning current will increase but after sometime, it becomes saturated.

Explanation:

Note: No information on change in number of electron generated.

Since there is a collision, the electrons emitted will not reach the collector at same time. As the voltage is increased, the the speed with which the electrons will reach the collector starts to increase. Due to this, electric current will first increases till all the emitted electrons reach the collector. Since we are not provided with the information that number of electrons generated are changing, after increasing voltage current will increase for some time and then reaches a saturated state.

We can conclude by saying that in the beginning current will increase but after sometime it becomes saturated.

4 0
3 years ago
What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co
MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0
3 years ago
A lever with an output arm of 0.8 meter has a mechanical advantage of 6. What is the length of the input
Novosadov [1.4K]
From the law of the lever: 

<span>mechanical advantage= input arm : output arm

input arm- a
output arm-b
</span>mechanical\ advantage= 6\\\\&#10;6=\frac{a}{0,8}\ \ |*0,8\\\\&#10;6*0,8=a\\\\&#10;a=4,8m&#10;&#10;
<span>Input arm is equal 4,8m</span>
4 0
4 years ago
Which type of rock can only form below earths surface?
Inessa [10]

Answer: Igneous it forms because of magma but magma is under the earths surface so its Igneous

Explanation:

7 0
2 years ago
Read 2 more answers
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
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