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MrRa [10]
3 years ago
6

Which of the following is equal to impulse? A. Ft B. m/s C. pt D. mv

Physics
1 answer:
tankabanditka [31]3 years ago
5 0
Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v
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A very long conducting cylinder (length L) of radius R(R<R and (b)r
mina [271]

Answer:

Explanation:

We have to find electric potential V at a distance r.

a) For r>R,

The electric field in the cylinder is given by

E.A equating it to the other electric field given by

б.A/ε₀

Here the area of cylinder is given by= 2*3.14*r*L

While for the outside, the area= 2*3.14*R*L

Equating both, we get

E= бR/rε₀

Now,

The potential difference is given as:

ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.

Where ΔV= V₀-V

So solving we get

V₀=V-бR/ε₀ln (r/R)

b) For r<R i.e. inside the cylinder

There will be no electric field produced as E=0

So ultimately Vin= V

c) V=0 at r= infinity.

4 0
3 years ago
Problems - Show all work.
Fittoniya [83]

Answer:

21s

Explanation:

Given parameters;

Radius  = 10m

Speed or velocity  = 3m/s

Unknown:

Period  = ?

Solution:

To solve this problem, use the expression:

      v  = \frac{2\pi r}{T}  

r is the radius

T is the unknown

           Input the parameters and solve for T;

    3  = \frac{2 x \pi  x 10}{T}  

    62.84 = 3T

         T  = 21s

4 0
3 years ago
A coil consists of 180 turns of wire. Each turn is a square of side d=30 cm, and a uniform magnetic field directed perpendicular
alexira [117]

Answer:

Emf induced i equal to 329.4 volt

Explanation:

Note : Here i think we have to find emf induced in the coil

Number of turns in the coil N= 180

Sides of square d = 30 cm = 0.3 m

So area of the square A=0.3\times 0.3=0.09m^2

Magnetic field is changes from 0 to 1.22 T

Therefore dB=1.22-0=1.22T

Time interval in changing the magnetic field dt = 0.06 sec

Induced emf is given by

e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

e=-180\times 0.09\times \frac{1.22}{0.06}=329.4volt

8 0
3 years ago
Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?
CaHeK987 [17]
In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.
7 0
3 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
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