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Paladinen [302]
3 years ago
6

A 30 kg student drops down from the monkey bars. The acceleration due to gravity is -9.8 m/s^2. Neglecting air drag, what is the

net force acting the on student?
A)-294 N
B)-3 N
C)0.3 N
D)294 N
Physics
1 answer:
Sholpan [36]3 years ago
8 0

The net force on the student is A) -294 N

Explanation:

Neglecting air resistance, there is only one force acting on the student: the force of gravity, which is given by

F=mg

where

m is the mass of the student

g is the acceleration of gravity

In this problem, we have:

m = 30 kg is the mass of the student

g=-9.8 m/s^2 is the acceleration of gravity, where the negative sign means the direction is downward

Substituting, we find the force of gravity on the student:

F=(30)(-9.8)=-294 N

And since this is the only force acting on the student, it is also the net force on him.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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Commercially-available hybrid vehicles, such as the Toyota Prius, use electrical batteries to store energy for later use. Howeve
Bad White [126]

(A) 4.2\cdot 10^5 J

The energy stored by the system is given by

E=Pt

where

P is the power provided

t is the time elapsed

In this case, we have

P = 60 kW = 60,000 W is the power

t = 7 is the time

Therefore, the energy stored by the system is

E=(60,000 W)(7 s)=4.2\cdot 10^5 J

(B) 4830 rad/s

The rotational energy of the wheel is given by

E=\frac{1}{2}I \omega^2 (1)

where

I is the moment of inertia

\omega is the angular velocity

The moment of inertia of the wheel is

I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2

where M is the mass and R the radius of the wheel.

We also know that the energy provided is

E=4.2\cdot 10^5 J

So we can rearrange eq.(1) to find the angular velocity:

\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s

(C) 2.8\cdot 10^6 m/s^2

The centripetal acceleration of a point on the edge is given by

a=\omega^2 R

where

\omega=4830 rad/s is the angular velocity

R = 0.12 m is the radius of the wheel

Substituting, we find

a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2

7 0
4 years ago
What is the period of a pendulum that takes 5 seconds to make a complete back and forth vibration
dezoksy [38]

Answer:

5s

Explanation:

-A period is the time it takes to make one full back-and-forth swing.

-A period is calculated as the time taken for a complete swing divided by one cycle:

T=t/1c\\\\=5s/1\\\\=5s

Hence, the period of the pendulum is 5s

8 0
4 years ago
At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291
Bogdan [553]

Answer:

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

8 0
3 years ago
An Olympic diver drops from the 10 meter platform with an initial velocity of 0.0m/s. What was the vertical velocity of the dive
sammy [17]

Answer:

v=14.14 m/s

t=1.141 s    

Explanation:

Given that

h = 10 m

Initial velocity ,u = 0 m/s

We know that acceleration due to gravity g= 10 m/s²

Lets take final velocity = v

Final velocity v is given as

v² = u²+ 2 g h

v²= 0² + 2 x 10 x 10

v²= 200

v=14.14 m/s

Time taken t is given as

v= u +  a t

a=acceleration

t=time

Now by putting the values in the above equation we get

14.14= 0 + 10 x t

14.14 = 10 t

t=1.141 s

4 0
4 years ago
What are the 7 electromagnetic waves in order from lowest to highest wavelength?
ozzi

The electromagnetic spectrum is divided into seven different frequency ranges which are from lowest to highest, radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays.

The electromagnetic spectrum includes electromagnetic waves with frequencies between one hertz and above 10²⁵ hertz, or wavelengths between thousands of kilometres and a small portion of the size of an atomic nucleus. The electromagnetic waves found within each of these bands are referred to by a different name. Starting at the low frequency(long wavelength), end of the spectrum, these are radio waves, micro waves, infrared, visible light, ultraviolet, x-rays and gamma rays. This frequency range is divided into separate bands.

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5 0
2 years ago
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