Ask question

Ask question

All categories

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

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

You might be interested in

What happens when charged object is brought near uncharged object?<br> Attract or Repel ?

Makovka662 [10]

When a charged object is brought near to but does not touch a neutral object, it causes the side of the neutral object that the charged object is near to become the other charge. It causes charge migration within the neutral object so the two charges (positive and negative) move to opposite sides of the object. Because the two objects do not touch, they do not repel each other, but rather have a slight attraction because of charge migration. If the two object were to touch then they would repel.

5 0

2 years ago

Read 2 more answers

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

3 years ago

How can you tell animal and plant cells apart

Lynna [10]

Well there are a couple ways. 1: The easiest, plant cells have a cell wall or extra protection. You will not find this on an animal cell, as there is only a membrane. 2: Chloroplast. This also will not be found in animal cells as they produce the plant's green color as well as the sugar. Hope this helps!

8 0

3 years ago

A model airplane of mass 0.6 kg is attached to a horizontal string and flies in a horizontal circle of radius 6 m, making 1.6 re

nikitadnepr [17]

Answer:

$speed= 12.15\frac{m}{s}$

Explanation:

In this question we have given

mass of airplane=.6 Kg

Radius of horizontal circle,r=6m

Time taken to complete 1.6 revolution=5s

Therefore time taken to complete 1 revolution, t= $\frac{5}{1.6}$

t=3.1 s

We have to find the speed of airplane

We will first find the distance covered by airplane in tracing one circle which is equal to circumference of circle

We know that

Circumference,d = $2 \pi \times r$

or $d=2\times 3.14 \times 6$

$d=37.68 m$

We know that speed= $\frac{d}{t}$ ............(1)

Put value of d and t in equation 1

speed= $\frac{37.68}{3.1}$

$speed= 12.15\frac{m}{s}$

5 0

3 years ago

Suppose that two sources of sound produce waves of the same exact amplitude and frequency, but are out of phase by one-half of a

tia_tia [17]

Answer:

option D is right

........

8 0

3 years ago

Read 2 more answers