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Blababa [14]
3 years ago
7

A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe

s.
What is the velocity of the ship in the direction of the current?
Physics
1 answer:
nataly862011 [7]3 years ago
7 0

Answer:

v_x = 11.78 m/s

Explanation:

Velocity of the ship is given as

v = 12 units

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

v_x = v cos\theta

v_x = 12 cos11

v_x = 11.78 m/s

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A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

\epsilon = -\frac{\Delta \Phi_B}{\Delta t} (1)

where

\Delta \Phi_B is the variation of magnetic flux through the coil

\Delta t = 2.0 ms = 0.002 s is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

\Phi_B = BA

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2

At the beginning, the magnetic field is

B_i = 1.5 T

so the flux is

\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

\Phi_f = 0

so the variation of magnetic flux is

\Delta \Phi = 0-0.068 Wb=-0.068 Wb

And substituting into (1) we find the average emf in the coil

\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0
3 years ago
A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equili
pentagon [3]

Answer:

x = 1.26 sin 3.16 t

Explanation:

Assume that the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

V=\dfrac{dx}{dt}

V= A ω cosωt

Maximum velocity

V(max)= Aω

Given that F= 32 N

F = K Δ

K=Spring constant

Δ = 0.4 m

32 =0.4 K

K = 80 N/m

We know that  ω²m = K

8 ω² = 80

ω = 3.16 s⁻¹

Given that V(max)= Aω = 4 m/s

3.16 A = 4

A= 1.26 m

Therefore the general equation of displacement

x = 1.26 sin 3.16 t

7 0
3 years ago
Read 2 more answers
radar wave is transmitted and later reflected off an aircraft and recieved 1.4×10^3 sec after being sent out. how far is the air
lutik1710 [3]

I think there's a typo because the answer I'm getting is very large.

This is what I'm getting

--------------------------------------

c = speed of light

c = 3.0 x 10^8 m/sec approximately

This is roughly 300 million meters per second

The time it takes the signal to reach the aircraft and come back is 1.4 x 10^3 seconds. Half of this time period is going one direction (say from the radar station to the aircraft), so (1.4 x 10^3)/2 = 7.0 x 10^2 seconds is spent going in this one direction.

distance = rate*time

d = r*t

d = (3.0 x 10^8) * (7.0 x 10^2)

d = (3.0*7.0) x (10^8*10^2)

d = 21.0 x 10^(8+2)

d = 21.0 x 10^10

d = (2.1 x 10^1) * 10^10

d = 2.1 x (10^1*10^10)

d = 2.1 x 10^11 meters

d = 210,000,000,000 meters (this is 210 billion meters; equivalent to roughly 130,487,950 miles)

3 0
3 years ago
A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0
4 years ago
When an ice cube in a glass of water melts, the water level _____. A. remains the same B. rises C. falls
tensa zangetsu [6.8K]
A.  The water level Remains the same
6 0
3 years ago
Read 2 more answers
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