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rusak2 [61]
3 years ago
15

A 92-kg receiver running at 8 m/s catches a pass across the middle and is immediately brought to a stop during a collision with

a safety that lasts 0.6 seconds. What was the magnitude of force needed to stop the receiver
Physics
1 answer:
sammy [17]3 years ago
7 0

Answer: 1200\ N

Explanation:

Given

Mass of receiver m=92\ kg

Running at a speed of v=8\ m/s

time taken to stop t=0.6\ s

We know, impulse imparted is given by

\Rightarrow F\cdot dt=m\Delta v\\\Rightarrow F(0.6)=90(8-0)\\\\\Rightarrow F=\dfrac{90\times 8}{0.6}\\\\\Rightarrow F=1200\ N

Thus, 1200 N is needed to stop the receiver.

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Ehren is trying to increase his mile run from eight to six minutes. He has decided to run some sandy hills near his home. Which
densk [106]

Answer:

The correct answer is intensity.

Explanation:

The principle of overload is a basic sports fitness training concept which means that in order to improve, athletes must continually work harder as they their bodies adjust to existing workouts. Thus, overloading also plays a role in skill learning.

Now, since erhen wants to increase his mile run from eight to six minutes by run some sandy hills near his home, it means the principle at work is making his training runs more intense for better speed results. Thus the principle at work is intensity because he is trying to do something that makes him run faster.

3 0
2 years ago
Help a Girl Out and Answer This you will get Brainliest, Thanks, and more points if your answer is correct (:
melamori03 [73]

Answer:

this was 2 weeks ago, but im pretty sure the correct answers are:

B=c

C=d

D=b

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3 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
2 years ago
NEED ASAP PLEASE !!
Svetllana [295]
The balloon was 30.65 meters above ground.

ANSWER: A
3 0
2 years ago
Read 2 more answers
A planet orbits a start with the path shown below.
kvv77 [185]

PART a)

As we know that gravitational potential energy is given by the formula

U = -\frac{Gm_1m_2}{r}

here we can see that gravitational potential energy inversely varies with the distance

so here when distance from the sun is minimum then magnitude of gravitational potential energy is maximum while since it is given with negative sign so its overall value is minimum at that position

So gravitational potential energy is minimum at the nearest point and maximum at the farthest point

PART b)

Since we know that sum of kinetic energy and potential energy is constant here

so the points of minimum potential energy is the point where kinetic energy is maximum which means speed is maximum

So here speed is maximum at the nearest point

Part C)

since gravitational potential energy inversely varies with distance so it's graph will be like hyperbolic graph with distance

4 0
3 years ago
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