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3 years ago

Similarities in tennis and volleyball

1 answer:

5 0

Volleyball and Tennis are similar in that there is a ball and a net involved in the games. In both games, you are facing the other team. in both games there is a method in scoring and up-er body strength involved.

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Does the refraction of light make a swimming pool seem deeper or shallower? explain your answer.

larisa [96]

The refraction of light makes a swimming pool seem shallower.

The swimming pool seems shallower because the rays of light coming from the bottom of the pool do not come with a straight path. The path of light is straight as long as it is in the water.

When lights come out of the water into the air it bents downwards. This bending is called refraction.

Refraction forms a virtual image of the pool and it seems shallower than it actually is to the observer. This only happens when light travels from one transparent medium into another having lower density.

If you need to learn more about why a swimming pool appears shallower, click here

https://brainly.in/question/7136803?referrer=searchResults

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6 0

1 year ago

The top of the pool table is 0.810 m from the floor. the placement of the tape is such that 0 m is aligned with the edge of the

8090 [49]

Compute first for the vertical motion, the formula is:

y = gt²/2

0.810 m = (9.81 m/s²)(t)²/2

t = 0.4064 s

whereas the horizontal motion is computed by:

x = (vx)t

4.65 m = (vx)(0.4064 s)

4.65 m/ 0.4064s = (vx)

(vx) = 11.44 m / s
So look for the final vertical speed.

(vy) = gt

(vy) = (9.81 m/s²)(0.4064 s)

(vy) = 3.99 m/s

speed with which it hit the ground:

v = sqrt[(vx)² + (vy)²]

v = sqrt[(11.44 m/s)² + (3.99 m/s)²]

v = 12.12 m / s

6 0

3 years ago

The periodic wave in the diagram below has a frequency of 80. hertz.

ASHA 777 [7]

Answer:

hi how are you

2) 5/8

4) 5/8

6 0

2 years ago

An object starts from rest at time t = 0.00 s and moves in the +x direction with constant acceleration. The object travels 14.0

Reptile [31]

Answer:

28 m/s^2

Explanation:

distance, s = 14 m

time, t = 2 - 1 = 1 s

initial velocity, u = 0 m/s

Let a be the acceleration.

Use third equation of motion

$s = ut + \frac{1}{2}at^{2}$

$14 = 0 + \frac{1}{2}a\times 1^{2}$

a = 28 m/s^2

Thus, the acceleration is 28 m/s^2.

7 0

3 years ago

Find the tension in the two ropes that are holding the 4.2 kg object in place.

Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0

3 years ago