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Galina-37 [17]
4 years ago
14

What is the most important

Physics
1 answer:
Andrew [12]4 years ago
4 0

Answer:

5

because while reading you should focus on your subjet when your mind will divert then you might reach to failure

You might be interested in
The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his
Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

F_x=kx

Substitute the values

6500=1300x

x=\frac{6500}{1300}=5m

Work done due to friction force

W_f=fscos\theta

We have \theta=180^{\circ}

Substitute the values

W_f=50\times 5cos180^{\circ}

W_f=-250J

Initial kinetic energy, Ki=0

Initial gravitational energy, U_{grav,1}=0\

Initial elastic potential energy

U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)

U_{el,1}=16250J

Final elastic energy,U_{el,2}=0

Final kinetic energy, K_f=\frac{1}{2}(60)v^2=30v^2

Final gravitational energy, U_{grav,2}=mgh=60\times 9.8\times 2.5

Final gravitational energy, U_{grav,2}=1470J

Using work-energy theorem

K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}

Substitute the values

0+0+16250-250=30v^2+1470+0

16000-1470=30v^2

14530=30v^2

v^2=\frac{14530}{30}

v=\sqrt{\frac{14530}{30}}

v=22m/s

5 0
3 years ago
PLEASE PROVIDE AN EXPLANATION.<br><br> THANKS!!!
ziro4ka [17]

Answer:

(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz

(b) 250 m/s

(c) 1250 N

(d) Positive x-direction

(e) 6.00 m/s

(f) 0.0365 m

Explanation:

(a) The standard form of the wave is:

y = A cos ((2πf) t ± (2π/λ) x)

where A is the amplitude, f is the frequency, and λ is the wavelength.

If the x term has a positive coefficient, the wave moves to the left.

If the x term has a negative coefficient, the wave moves to the right.

Therefore:

A = 0.0800 m

2π/λ = 0.300 m⁻¹

λ = 20.9 m

2πf = 75.0 rad/s

f = 11.9 Hz

(b) Velocity is wavelength times frequency.

v = λf

v = (20.9 m) (11.9 Hz)

v = 250 m/s

(c) The tension is:

T = v²ρ

where ρ is the mass per unit length.

T = (250 m/s)² (0.0200 kg/m)

T = 1250 N

(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).

(e) The maximum transverse speed is Aω.

(0.0800 m) (75.0 rad/s)

6.00 m/s

(f) Plug in the values and find y.

y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))

y = 0.0365 m

8 0
3 years ago
Read 2 more answers
Determine the initial velocity of the ball if it reaches a height of 15 meters.
jasenka [17]

Answer:

the initial velocity of the  ball is 17.14 m/s

Explanation:

Given;

maximum height reached by the ball, h = 15 m

let the initial velocity of the ball = u

at maximum height, the final velocity of the ball is zero, v = 0

The initial velocity of the ball is calculated by using the following upward motion kinematic equation;

v² = u² - 2gh

0 = u² - 2(9.8 x 15)

u² = 294

u = √294

u = 17.14 m/s

Therefore, the initial velocity of the  ball is 17.14 m/s

4 0
3 years ago
What force is applied to a 60 kg person if it takes 7.8 seconds to reach a speed of 12.86 m/s from rest?
docker41 [41]

Answer:

drag

Explanation:

drag is the force that acts against any moving object.

if it takes him any amount of time to get from one speed to another, it is caused by drag.

3 0
2 years ago
The tape in a videotape cassette has a total
xxMikexx [17]

Answer:

w=19.76 \ rad/s

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

w=w_o+\alpha \ t

Where \alpha is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

\displaystyle v_t=w.r

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call w_1 the angular speed of the loaded reel and w_2 that from the empty reel. We have

w_1=w_{o1}+\alpha_1 \ t

w_2=w_{o2}+\alpha_2 \ t

If r_f=35\ mm=0.035\ m is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

\displaystyle w_{01}=\frac{v_t}{r_f}

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

t=1.8\ h=1.8*3600=6480\ sec

\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s

\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s

If r_e=10\ mm=0.001\ m is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s

The full reel goes from w_{01} to w_{02} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}

\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2

The empty reel goes from w_{02} to w_{01} in 6480 seconds, so we can compute the angular acceleration:

\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2

So the equations for both reels are

w_1=1.098+0.00576 \ t

w_2=38.426-0.00576 \ t

They will be the same when

1.098+0.00576 \ t=38.426-0.00576 \ t

Solving for t

\displaystyle t=\frac{38.426-1.098}{0.0115}

t=3240 \ sec

The common angular speed is

w_1=1.098+0.00576 \ 3240=19.76 \ rad/s

w_2=38.426-0.00576 \ 3240=19.76 \ rad/s

They both result in the same, as expected

\boxed{w=19.76 \ rad/s}

6 0
4 years ago
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