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myrzilka [38]
3 years ago
14

The land between two normal faults moves upward to form a

Physics
2 answers:
ahrayia [7]3 years ago
6 0

THE ANSWER IS D .................................

Leya [2.2K]3 years ago
5 0
<span>The land between two normal faults moves upward to form a

Answer:D</span><span>
fault-block mountain.</span>
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A man paddles a canoe in a long, straight section of a river. The canoe moves downstream with constant speed 5 m/s relative to t
nikklg [1K]

Answer:

Explanation:

We shall consider all movement with respect to water assuming that river is at rest or motionless .

speed of canoe = 5 m /s

in five minutes , distance between hat and canoe = 5 x 60 x 5

1500 m

This distance will be covered by man in return journey . In this case his speed is 5 m /s again considering river constant .

So this will be covered at 5 m /s

time taken = 1500 / 5 = 300 s

= 5 minutes .

so it will take 5 minutes to row back to reclaim his hat .

7 0
3 years ago
1- Show that the equation d=vt2<br> dimensionally correct or not?
TiliK225 [7]
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5 0
2 years ago
Infrared radiation _____. has wavelengths that are shorter than visible light is part of the visible light spectrum has waveleng
iren [92.7K]

Answer: has wavelengths that are longer than visible light

Explanation: Electromagnetic spectra comprises of different electromagnetic waves which are arranged in order of increasing wavelengths or decreasing energies.

In order from highest to lowest energy, the waves are arranged as: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves.

E=\frac{hc}{\lambda}

E= energy

h= planck's constant

c= velocity of light

\lambda = wavelength

As wavelength and energy are inversely related, the wave having higher wavelengths will be lowest in energy. Thus as infrared radiation have lower energy as compared to visible radiation. Thus  infrared radiation have higher wavelength as compared to visible radiation.

5 0
3 years ago
When a wire 1.5 m long carries a 24-A current in the +x direction in a uniform external magnetic field, the magnetic force exert
VLD [36.1K]

Answer:

Explanation:

Let the magnetic field be

B = B_xi +B_yj +B_z k

For magnetic force , the expression is

F = L ( I x B )

= 1.5 ( 24i x B_xi +B_yj +B_z k )

F = 36 B_y k - 36 B_z j

Given

F = 3 j + 2 k

Equating equal terms

we have

B_y = 2 / 36 , B_z = - 3 / 36

Now the direction of current is changed to y direction

so F = 1.5[ 24j \times ( B_xi +B_yj +B_z k )]

Given

F = - 3 i - 2 k

Equating equal terms

we have

B_x = 2 / 36 , B_z = - 3 / 36

So B = 2/36 i + 2/36 j - 3/ 36 k

Magnitude of B

= 4.1 / 36 T

8 0
3 years ago
At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until
Sav [38]

Answer:

a) \Delta \theta = 617.604\,rad, b) \Delta t = 10.392\,s, c) \alpha = -14.128\,\frac{rad}{s^{2}}

Explanation:

a) The final angular speed at the end of the acceleration stage is:

\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)

\omega = 111.5\,\frac{rad}{s}

The angular deceleration is:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}

\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}

\alpha = -14.128\,\frac{rad}{s^{2}}

The change in angular position during the acceleration stage is:

\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}

\theta = 177.604\,rad

Finally, the total change in angular position is:

\Delta \theta = 440\,rad + 177.604\,rad

\Delta \theta = 617.604\,rad

b) The time interval of the deceleration interval is:

\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }

\Delta t = 7.892\,s

The time required for the grinding wheel to stop is:

\Delta t = 2.50\,s + 7.892\,s

\Delta t = 10.392\,s

c) The angular deceleration of the grinding wheel is:

\alpha = -14.128\,\frac{rad}{s^{2}}

4 0
3 years ago
Read 2 more answers
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