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n200080 [17]
2 years ago
11

Explain how the basic units are combined to give the derived units of force, velocity and pressure and work​

Physics
1 answer:
Lorico [155]2 years ago
5 0

#Force

\\ \sf\longmapsto Force=Mass\times Acceleration

\\ \sf\longmapsto Force=kg m/s^2

\\ \sf\longmapsto Force=Newton (N)

#Velocity

\\ \sf\longmapsto Velocity=\dfrac{Displacement}{time}

\\ \sf\longmapsto Velocity=m/s

#Work

\\ \sf\longmapsto Work=Force\times Displacement

\\ \sf\longmapsto Work=Nm

\\ \sf\longmapsto Work=Joules(J)

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Suppose 2.10 C of positive charge is distributed evenly throughout a sphere of 1.30-cm radius. 1) What is the charge per unit vo
Anuta_ua [19.1K]

Answer:

\rho=2.28\times 10^5\ C/m^3

Explanation:

Given that,

Charge, Q = 2.1 C

The radius of sphere, r = 1.3 cm = 0.013 m

We need to find the charge per unit volume for this situation. It can be calculated a follows:

\rho=\dfrac{Q}{\dfrac{4}{3}\pi r^3}\\\\\rho=\dfrac{2.1}{\dfrac{4}{3}\pi \times (0.013)^3}\\\\\rho=2.28\times 10^5\ C/m^3

So, the charge per unit volume is 2.28\times 10^5\ C/m^3.

3 0
2 years ago
A stone of weight 10N falls from the top of a 250m high cliff. a) Calculate how much work is done by the force of gravity in pul
tiny-mole [99]

Answer:

work done = ( force × displacement)

(a)The force acting on the block is it's self weight and displacement is equal to height of the tower.

work done by gravity = (250 × 10) = 2500 joule

(b) The work done by gravity 2500 joule is transferred to the object in the form of it's kinetic energy.

4 0
1 year ago
According to Archimedes’ principle, the mass of a floating object equals the mass of the fluid displaced by the object. Use this
Andrew [12]

Answer:

Part a)

\rho = 0.55 g/cm^3

Part b)

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

Explanation:

Part a)

As we know that when cylinder float in the water then weight of the cylinder is counter balanced by the buoyancy force

So here we know

buoyancy force is given as

F_b = \rho_w V_{sub} g

F_b = (1 g/cm^3) (30 - 13.5) Ag

F_b = 16.5 Ag

Now we know that the weight of the cylinder is given as

W = \rho (30 cm)A g

now we have

\rho (30 cm) A g = 16.5 A g

\rho = 0.55 g/cm^3

Part b)

When the same cylinder is floating in other liquid then we will have

F_b = \rho_L (30 - 18.9 )A g

so we have

\rho_L (11.1) Ag = 0.55(30) Ag

\rho_L = 1.49 g/cm^3

Part c)

Since we know that the base area will remain same always

so here the length and width of the object is not necessary to obtain the above data in such type of questions

3 0
2 years ago
Assuming 70% of Earth's surface is covered in water at an average depth of 2.5 mi, estimate the mass of the water on Earth in Ki
saw5 [17]
This is an excellent question that i do not have the answer to.
7 0
2 years ago
A wall clock has a minute hand with a length of 0.55 m and an hour hand with a length of 0.26 m. Take the center of the clock as
Lemur [1.5K]

Answer:

The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

Explanation:

Given that,

Length of minute hand = 0.55 m

Length of hour hand = 0.26 m

The time taken by the minute hand to complete one revelation is

T= 3600\ sec

We need to calculate the angular frequency

Using formula of angular frequency

\omega=\dfrac{2\pi}{T}

Put the value into the formula

\omega=\dfrac{2\pi}{3600}

\omega=0.001745\ rad/s

We need to calculate the magnitude of the acceleration of the tip of the minute hand of the clock

Using formula of acceleration

a=r\omega^2

Put the value into the formula

a=0.55\times(0.001745)^2

a=1.675\times10^{-6}\ m/s^2

Hence, The magnitude of the acceleration of the tip of the minute hand of the clock 1.675\times10^{-6}\ m/s^2.

3 0
2 years ago
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