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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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lbvjy [14]

The new current will be 4I. The power dissipated by the resistor will increase by a factor of 16.

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4 0
1 year ago
A ball rolls downhill with a constant acceleration of 4m/s squared. If it started from rest,it’s velocity at the end of 3 meters
vladimir2022 [97]

Answer:

4.9 m/s

Explanation:

Since the motion of the ball is a uniformly accelerated motion (constant acceleration), we can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the ball in this problem,

u = 0 (it starts from rest)

a=4 m/s^2 is the acceleration

s = 3 m is the distance covered

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(4)(3)}=4.9 m/s

3 0
4 years ago
A proton is 0.9 meters away from a 1.4 C charge. What is the magnitude of the electric force between the proton and the charge
Digiron [165]

Answer:

F = 2.49 x 10⁻⁹ N

Explanation:

The electrostatic force between two charged bodies is given by Colomb's Law:

F = \frac{kq_1q_2}{r^2}\\

where,

F = Electrostatic Force = ?

k = colomb's constant = 9 x 10⁹ N.m²/C²

q₁ = charge on proton = 1.6 x 10⁻¹⁹ C

q₂ = second charge = 1.4 C

r = distace between charges = 0.9 m

Therefore,

F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(1.6\ x\ 10^{-19}\ C)(1.4\ C)}{(0.9\ m)^2}

<u>F = 2.49 x 10⁻⁹ N</u>

8 0
3 years ago
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Rock at the top of a 20 m tall hill the rock has a mass of 10 kg how much potential energy does it have
Blizzard [7]
PE = 10 * 10 * 20 = 2000 Joule
6 0
3 years ago
Read 2 more answers
3.00 m^3 of water is at 20.0°C.
krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume  v1=  3m^3

volume  v2=  ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

6 0
3 years ago
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