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g100num [7]
1 year ago
7

Diego is playing basketball. While running at 7 km/h toward Gino, he passes the ball to Gino horizontally. The ball travels at 2

0 km/h relative to Diego.
Physics
1 answer:
enot [183]1 year ago
3 0

TheThe  relative velocity of Deigo will be 27km/h .

<h3>What is relative Velocity? </h3>

The relative velocity is the velocity of an object with respect to another observer. It is the time rate of relative position of one object with respect to another object .

Vab = Va-Vb

Where a is object and b is observer .

We have given here the velocity of ball 20km/ h

and the another velocity is 7km/ h

Let us assume that Diego is also moving in the direction where ball is moving so the relative velocity  will  be

 V = 20+7=27km/ h

When both observer and object is moving in same direction than there sum of velocity will be relative velocity ,if they are moving opposite than subtraction.

to learn more about Relative velocity click Here brainly.com/question/6070547

#SPJ4

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Isaac Newton used a ____ to break white light into its component colors.
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<h2>Answer: Prism</h2>

In the eighteenth century Isaac Newton found out that <u>when a beam of light from the Sun, passes trhough a prism is decomposed in many different colors</u>. He named this phenomenom as dispersion of light.

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Therefore:

<h2>Isaac Newton used a <u>prism</u> to break white light into its component colors.</h2>
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Describe how to make a position-time graph.
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3 0
3 years ago
A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees. What is the vertical component of t
Svetllana [295]

If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees

The horizontal component of the velocity is given by

Vx = Vcosθ

The vertical component of the velocity is given by

Vy = Vsinθ

As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground

Vy = 16.71 × sin49.21°

Vy = 12.65 m/s

Thus, the vertical component of the velocity would be 12.65 m/s

Learn more about Velocity from here

brainly.com/question/18084516

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4 0
2 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0
3 years ago
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