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Lerok [7]
3 years ago
7

State three factors on which the pressure at a point in a liquid depends.

Physics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

???

Explanation:

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Why is it possible for a gas to take the shape of its container?
Sedaia [141]
I believe the answer is A

8 0
2 years ago
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Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive
Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

r=\dfrac{d}{2}

r=\dfrac{15\times9.461\times10^{15}}{2}

r=7.09\times10^{16}\ m

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

M=\dfrac{v^2r}{G}

Put the value into the formula

M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}

M=3.44\times10^{37}\ Kg

Hence, The mass of the massive object at the center of the Milky Way galaxy is 3.44\times10^{37}\ Kg

5 0
3 years ago
The area of the large syringe in an experiment is 18cm2 and that of the smaller one is 3.0cm2. A force of 2N is applied on the s
lilavasa [31]

Explanation:

F/18=2/3

F=2×18/3

F=12N

3 0
3 years ago
The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil
gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section=3.7\times 3.7 cm^2=13.69 cm^2

N=no of turns=61

Resistance=R=8 \Omega

current (i)=0.04 A

emf induced=i\times R=0.04\times 8=0.32 V

emf induced e=NA\frac{\mathrm{d} B}{\mathrm{d} t}

0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}

\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s

8 0
3 years ago
A spring stretches 3.5 cm when a 9 g object is hung from it. The object is replaced with a block of mass 26 g which oscillates i
evablogger [386]

Answer:

The period of motion  of new mass T = 0.637 sec

Explanation:

Given data

Mass of object (m) = 9 gm = 0.009 kg

Δx = 3.5 cm = 0.035 m

We know that spring force is given by

F = m g

F = 0.009 × 9.81 = 0.08829 N

Spring constant

k = \frac{F}{x}

k = \frac{0.08829}{0.035}

k = 2.522 \frac{N}{m}

New mass(m_1) = 26 gm = 0.026 kg

Now the period of motion is given by

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

T = 2 \pi \sqrt{\frac{0.026}{2.522} }

T = 0.637 sec

This is the period of motion  of new mass.

3 0
2 years ago
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