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3 years ago

6

What's the volume of the tennis ball

2 answers:

MariettaO [177]3 years ago

7 0

It is eight cubic inches

Lubov Fominskaja [6]3 years ago

3 0

8 cubic inches is the volume of a tennis ball

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Properties of convex mirror

Ahat [919]

Answer:

The image result of an object reflected by a convex mirror is typically virtual, upright, and smaller. Discover how moving the object farther away from the mirror's surface affects the size of the virtual image formed behind the mirror

Explanation:

6 0

2 years ago

The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

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3 years ago

Which wave has a disturbance that is parallel to the wave motion?

Leto [7]

The answer is D.<span>longitudinal</span>

6 0

2 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

A gymnast of mass 63.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

Sergio [31]

Answer:

Explanation:

A ) When gymnast is motionless , he is in equilibrium

T = mg

= 63 x 9.81

= 618.03 N

B )

When gymnast climbs up at a constant rate , he is still in equilibrium ie net force acting on it is zero as acceleration is zero.

T = mg

= 618.03 N

C ) If the gymnast climbs up the rope with an upward acceleration of magnitude 0.600 m/s2

Net force on it = T - mg , acting in upward direction

T - mg = m a

T = mg + m a

= m ( g + a )

= 63 ( 9.81 + .6)

= 655.83 N

D ) If the gymnast slides down the rope with a downward acceleration of magnitude 0.600 m/s2

Net force acting in downward direction

mg - T = ma

T = m ( g - a )

= 63 x ( 9.81 - .6 )

= 580.23 N

6 0

2 years ago