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Triss [41]
3 years ago
14

A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a

t 4.2 m/s when he was tackled by an 85 kg cornerback running at 5.5 m/s in the same direction, what was their mutual speed immediately after the touchdown-saving tackle? m/s
Physics
1 answer:
uranmaximum [27]3 years ago
3 0

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}

Where, m_{h}= mass of halfback

m_{c}=mass of corner back

v_{h}= velocity of halfback

v_{c}= velocity of corner back

Put the value into the formula

98\times4.2+85\times5.5=(98+85)\times v

v=\dfrac{98\times4.2+85\times5.5}{98+85}

v=4.80\ m/s

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

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Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s

 

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}

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3 years ago
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o
Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

#SPJ1

8 0
2 years ago
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podryga [215]

Answer:

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3 years ago
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2. Which of the following is NOT true of work?
salantis [7]

Answer:

D

Explanation:

Work is not a vector but it is a scalar

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3 years ago
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Only 35 % of the intensity of a polarized light wave passes through a polarizing filter. What is the angle between the electric
Nana76 [90]

Answer:

The angle between the electric field and the axis of the filter is 54⁰

Explanation:

Apply the equation for intensity of light through a polarizer.

I = I_oCos^2 \theta

where;

I is the intensity of the transmitted light

I₀ is the intensity of the incident light

θ is the incident angle

If only 35 % of the intensity of a polarized light wave passes through a polarizing filter, then the ratio of the intensity of the transmitted light to that of the intensity of the incident light is given by;

\frac{I}{I_o}  = Cos^2 \theta\\\\\frac{35}{100} =  Cos^2 \theta\\\\Cos^2 \theta = 0.35\\\\Cos\theta = \sqrt{0.35} \\\\Cos\theta = 0.5916\\\\\theta = Cos^{-1}(0.5916)\\\\\theta  = 54 ^0

Therefore, the angle between the electric field and the axis of the filter is 54⁰

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3 years ago
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