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Sliva [168]
3 years ago
13

A ball is thrown horizontally from a high cliff with a velocity of 5 m/s. Ignoring the slowing of the ball due to air resistance

(and assuming the ball is still falling), what will be the horizontal velocity of the ball 3 seconds later? A. 0 m/s B. 5 m/s C. 9.8 m/s D. 15 m/s
Physics
2 answers:
Aleksandr-060686 [28]3 years ago
7 0
If there's no air resistance, then there are no horizontal forces on the ball.
With no horizontal force, there's no horizontal acceleration, so the horizontal speed never changes.  From the time it's tossed until it hits the ground, its horizontal speed is the same 5 m/s. (B)
viktelen [127]3 years ago
5 0
Hello 
my dude 
how are u 

so the answer to this is going to be 

B

hope this helps 

have a good day
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For a steady two-dimensional flow, identify the boundary layer approximations.
Georgia [21]

Answer:

  • The velocity component in the flow direction is much larger than that in the normal direction ( A )
  • The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )

Explanation:

For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction

assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that

Vy << Vx

4 0
3 years ago
Consider an object with weight on the Earth W_earth. The gravity of the Earth is g. If it is moved to another planet with gravit
Mazyrski [523]

Answer:

W_{Planet} = \frac{W_{Earth}}{g}\times a

Explanation:

We know that weight of an object on Earth is,

W_{Earth} = m\times g

Thus,

m = \frac{W_{Earth}}{g}

where,

m = mass of an object, which is constant and is independent of gravity

g = acceleration due to gravity on Earth

On the new planet, gravity = a

Thus the weight of the object on the new planet will be

W_{Planet} = m\times a

W_{Planet} = \frac{W_{Earth}}{g}\times a

5 0
2 years ago
Explain why Earth is not spherical in shape, but bulges at<br> theEquator?
Strike441 [17]

Explanation:

Every rotating body experiences centrifugal force. Due to this force the body tends to bulge out around it mid point and gets flattened at the poles. Same is applicable to Earth as well. Since the Earth is rotating at a very high speed, its equator gets bulged out due to centrifugal force. Because of this bulged equator, Earth's pole to pole diameter and equatorial diameter has difference of around 42.76 km. It is flatter on the poles. This also proves that Earth is not a perfect sphere.

8 0
2 years ago
Read 2 more answers
The tungsten filament of a light bulb has an operating temperature of about 2 100 K. If the emitting area of the filament is 1.0
Murljashka [212]

Answer:

75 W

Explanation:

T = temperature of the filament = 2100 K

A = Emitting area of the filament = 1 cm² = 10⁻⁴ m²

e = Emissivity = 0.68

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻²K⁻⁴

Using Stefan's law, Power output of the light bulb is given as

P = \sigma e AT^{4} \\P = (5.67\times10^{-8}) (0.68) (10^{-4}) (2100)^{4}\\P = 75 W

5 0
2 years ago
two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times
Marina CMI [18]

Explanation:

The initial kinetic energy KE_0 for both blocks is zero. Let m_1= m and m_2 =3m. So using the conservation law of linear momentum, we can write

0 = m_1v_1 - m_2v_2

or

v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1

The final kinetic energies for the two masses are

KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2

KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)

Therefore, the ratio of their kinetic energies is

\dfrac{\Delta KE_2}{\Delta KE_1}  = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}

4 0
2 years ago
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