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Fittoniya [83]
2 years ago
15

he electric motor of a model train accelerates the train from rest to 0.720 m/s in 29.0 ms. The total mass of the train is 840 g

. Find the average power delivered to the train during its acceleration. W
Physics
1 answer:
kkurt [141]2 years ago
4 0

Answer:

7.481W

Explanation:

power= work done÷time

power= force × uniform velocity

a= ∆v/t

29.0ms=

29 \times  {10}^{ - 3 } s

=(0.72-0)÷29×10^-3

a=24.83m/s^2

Newton's second law

f=ma

=(0.84×24.83)

=20.86N

w=FD

v^2=u^2+2as

0.5184=0+49.66s

s=0.0104m

20.86×0.0104

=0.216944J

0.216944÷t

P=7.481W

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A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star
LiRa [457]

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

8 0
2 years ago
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The
nordsb [41]

Answer:

t=6.96s

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

v_{oy}=96ft/s

y_{o}=112ft

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}

0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}

Solving for t using quadratic formula

t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}

a=-\frac{1}{2} (32.2)\\b=96\\c=112

t=-0.999s or t=6.96s

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0
3 years ago
A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w
Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

\theta= \omega t

and substituting t = 75 seconds, we find

\theta= (0.20 rad/s)(75 s)=15 rad

In degrees, it is

15 rad: x = 2\pi rad : 360^{\circ}\\&#10;x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
For the airfoil and conditions in Problem 2.2, calculate the lift-to-drag ratio. Comment on its magnitude.
raketka [301]

Answer:

L/D= 112

Explanation:

Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.

Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.

Lift increases proportionally to the square of the speed.

The solutions to the question is the file attached to this explanation.

Lift,L= qC(l). S---------------------------(1).

and,

Drag,D = qC(d).S ----------------------(2).

Hence, Lift to drag ratio,L/D= C(l)/C(d).

Therefore, we have to compute various angle of attack.(check attached file)...

Then, (L/D) will then be equal to 112.

8 0
3 years ago
A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o
Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}.

Kinetic energy of system at first: \frac{mv^2}{2}=784\;\textbf{J}. After: \frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}. The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0
3 years ago
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