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dem82 [27]
4 years ago
9

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

ion of 4.5 m/s squared
Physics
1 answer:
Gekata [30.6K]4 years ago
5 0

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}

\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}

\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}

Using Newton's second law, it is established that  F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object \left(m_{1}+m_{2}\right)

Now consider both \mathrm{m}_{1} \text { and } \mathrm{m}_{2}as a system, so net force acting on the system is \text { Force }=\left(m_{1}+m_{2}\right) a

Substitute the given values in the above formula,

\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

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The volume flow rate of blood leaving the heart to circulate throughout the body is about 5 L/min for a person at rest. All this
skelet666 [1.2K]

Answer:

n=2.9\times 10^9

A=1.88\times 10^{-8}\ m^2

Explanation:

Given that

Q= 5 L/min

1 L = 10⁻³ m³/s

1 min = 60 s

Q=0.083 x 10⁻³ m³/s

d= 6 μm

v= 1 mm/s

So the discharge flow through one tube

q = A v

A=\dfrac{\pi}{4}d^2

A=\dfrac{\pi}{4}\times (6\times 10^{-6})^2\ m^2

A=2.8 x 10⁻¹¹ m²

v= 1 x 10⁻³  m/s

q= 2.8 x 10⁻¹⁴  m³/s

Lets take total number of tube is n

Q= n q

n=Q/q

n=\dfrac{0.083\times 10^{-3} }{ 2.8\times 10^{-14}}

n=2.9\times 10^9

Surface  area A

A= π d L

A=\pi \times 6\times 10^{-6}\times 10^{-3}\ m^2

A=1.88\times 10^{-8}\ m^2

7 0
3 years ago
Your little sister is building a radio from scratch. Plans call for a 500 μH inductor wound on a cardboard tube. She brings you
gayaneshka [121]

Answer:

N = 195 turns

Explanation:

The inductance of the inductor, L = 500 μH = 500 * 10⁻⁶H

The length of the tube, l = 12 cm = 0.12 m

The diameter of the tube, d = 4 cm = 0.04 m

Radius, r = 0.04/2 = 0.02 m

Area of the tube, A = πr² = 0.02²π = 0.0004π m²

\mu_{0} = 4\pi * 10^{-7}

The inductance of a solenoid is given by:

L = \frac{\mu_{0}N^{2} A }{l}

500 * 10^{-6} = \frac{4\pi *10^{-7}  N^{2} *4\pi  *10^{-4}  }{0.12}\\500 * 10^{-6} = 0.00000001316N^{2} \\N^{2} = \frac{500 * 10^{-6}}{0.00000001316}\\N^{2} = 37995.44\\N = \sqrt{37995.44} \\N = 194.92 turns

8 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
xxMikexx [17]
14-6 =8
8/4= 2m/s per second
8 0
3 years ago
Read 3 more answers
Which of the following is an abiotic limiting factor for a plant population in an ecosystem?
Ludmilka [50]

Correct answer is Availability of soil minerals.

There are two types of limiting factors that affect plant population in a ecosystem. They are:

  1. Biotic
  2. Abiotic

Biotic factors includes food, diseases etc. Such as Invasive weed species, seed dispersal by wind, disease-causing fungal spores are examples of biotic limiting factor.

Abiotic factors include sunlight, temperature and chemical environment. The availability of soil mineral is an example of abiotic limiting factor. Growth of plant population depends on availability of soil minerals.

8 0
3 years ago
What is the net force on this object f air = 400n (up) fgrav=600n (down)
n200080 [17]
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N
5 0
3 years ago
Read 2 more answers
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