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dem82 [27]
3 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]3 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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Four forces act on a hot-air balloon, as shown
dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

  • <em>Upward force = 514 N</em>
  • <em>Downward force = 267 N</em>
  • <em>Eastward force = 678 N</em>
  • <em>Westward force = 397 N</em>

The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

The net horizontal force on the balloon is calculated as follows;

F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

The magnitude of the resultant force on the balloon is calculated as follows;

F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0
2 years ago
Which of the following describes a referee's job?
Serhud [2]

Answer:

C. Supervising the game to make sure teams are playing fairly

5 0
4 years ago
Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t
Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector =7cos30^{\circ}=7\times 0.866=6.06

y component of the vector =7sin30^{\circ}=7\times 0.5=3.5

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector  =7cos120^{\circ}=7\times -0.5=-3.5i

y component of the vector =7sin120^{\circ}=7\times 0.866=6.06j

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0
3 years ago
7. A car moving at 10m/s (about 22.4 mph) crashes into a barrier and stops in 0.25 m.
Galina-37 [17]

Answer:

a) 0.05s

b) 4000N

Explanation:

a)When car is stopped its final velocity become zero

U- 10 m/s

V- 0 m/s

S - 0.25 m

t -?

S = (v+u)*t/2

0.25 =(10+0)*t/2

t = 0.05s

b) If we happened to calculate the avarage force we have to consider about acceleration

V= 0

U = 10

t = 0.05 s

a =?

V = U + at

0 = 10 -a * 0.05

a = 200 m/s2

F = m *a

= 20 * 200

= 4000N

6 0
3 years ago
If an object accelerates from rest, with a constant of 8 m/s2, what will its velocity be after 35s?
AlladinOne [14]

Answer:

<em>Its speed will be 280 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.

If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

v_f=v_o+at

The object accelerates from rest (vo=0) at a constant acceleration of a=8\ m/s^2. The final speed at t=35 seconds is:

v_f=0+8*35

v_f=280\ m/s

Its speed will be 280 m/s

5 0
3 years ago
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