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lapo4ka [179]
3 years ago
11

To become a personal trainer, a bachelor's degree is usually needed.

Physics
2 answers:
olya-2409 [2.1K]3 years ago
7 0
The Correct Answer Is False.
Eddi Din [679]3 years ago
4 0
<span>It is false that in order to become a personal trainer, a bachelor's degree is usually needed. The minimal requirements for such a job would be a diploma, or a certificate that you have the necessary skills and knowledge in order to become a personal trainer, but you do not need a faculty degree in order to perform this job. Your skills are far more improtant than the paper from a university.</span>
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A race car starting from rest accelerates uniformly at 4.9 m/s^2. What is the car's speed after it has traveled 200 meters?
NeX [460]
The equation v²=u²+2as will help with solving this equation
v is what we are trying to find
u=0
a=4.9
s=200
v²=0²+2*4.9*200
v=√(2*4.9*200)
v=√1960
v=44.3 m/s (3 sf)
5 0
4 years ago
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Find the current through a loop needed to create a maximum torque of 9.00 mN. The loop has 50 square turns that are 15.0 cm on a
Umnica [9.8K]

Answer:

10 A

Explanation:

τ = Maximum torque of the loop = 9 mN

N = Number of turns in the loop = 50

a = side of the loop = 15 cm = 0.15 m

A = Area of the loop = a² = 0.15² = 0.0225 m²

B = magnitude of magnetic field = 0.800 T

i = magnitude of current in the loop

Maximum torque of the loop is given as

τ = N B i A

Inserting the values

9 = (50) (0.8) (0.0225) i

i = 10 A

6 0
3 years ago
What part of earth systems interact to form a storm like this hurricane near Florida
Akimi4 [234]
Hurricanes form from interactions between the atmosphere and the oceans. Hope it helps.
7 0
3 years ago
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
A scientist studies a model of a widely accepted theory about the position of planets and sun. The model would most likely help
Troyanec [42]
If a scientist is studying a model that is a widely accepted theory about the position of planets and sun, the model would most likely help the scientist to understand facts that can't be easily observed. That will make the correct answer to be C. 
4 0
3 years ago
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