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Law Incorporation [45]
3 years ago
5

A train accelerates from rest and covers a distance of 600m in 20s. What is the train's speed at the end of 20s? What is the fin

al velocity of the train after 20s?
Physics
1 answer:
Allushta [10]3 years ago
6 0

Answer:

v = 60 m/s

final velocity after 20 seconds depends on how long it continues to accelerate after 20 seconds. Also we do not know the direction to complete the velocity requirements of a scalar value and direction component.

Explanation:

If we assume that the acceleration is constant

d = ½at² = ½(Δv/t)t² = ½Δvt = ½(v - 0)t = ½vt

v = 2d/t

v = 2(600)/20 = 60m/s

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A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the
zhannawk [14.2K]

Answer:

1.429*10^-5 m

Explanation:

From the question, we are given that

Diameter of the cable, d = 3 cm = 0.03 m

Force on the cable, F = 2 kN

Young Modulus, Y = 2*10^11 Pa

Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²

The fractional length = Δl/l

Δl/l = F/AY

Δl/l = 2000 / 0.0007 * 2*10^11

Δl/l = 2000 / 1.4*10^8

Δl/l = 1.429*10^-5 m

Therefore, the fractional length is 1.429*10^-5 m long

6 0
3 years ago
um can someone please help me I'm really stuck on this question and an explanation would be nice thank you
masha68 [24]

Answer: A wavelength is how long ONE wave is. First divide the distance of the whole diagram by the number of waves. you will get 2m. This is the answer.

6 0
2 years ago
An electron and a proton are both released from rest, midway between the plates of a charged parallel-plate capacitor. The only
topjm [15]

Answer:

Explanation:

Let the potential difference between the middle point and one of the plate be ΔV .

electric potential energy will be lost and it will be converted into kinetic energy .

Electrical potential energy lost = Vq , where q is charge on charge particle .

For proton

ΔV× q = 1/2 M V² ( kinetic energy of proton )

where M is mass and V be final velocity of proton .

For electron

ΔV× q = 1/2 m v² ( kinetic energy of electron  )

where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .

As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same

1/2 M V² =  1/2 m v²

(V / v )² = ( m / M )

(V / v ) = √ ( m / M )

In other words , their velocities  are  inversely proportional to square root of their masses .

4 0
3 years ago
A pitcher delivers a fast ball with a velocity of 43 m/s to the south. the batter hits the ball and gives it a velocity of 51 m/
hoa [83]
Assuming north as positive direction, the initial and final velocities of the ball are:
v_i=-43 m/s (with negative sign since it is due south)
v_f=+51 m/s
the time taken is t=1.0 ms=0.001 s, so the average acceleration of the ball is given by
a= \frac{v_f-v_i}{t}= \frac{51 m/s-(-43 m/s)}{0.001 s}=9.4 \cdot 10^4 m/s^2
And the positive sign tells us the direction of the acceleration is north.
4 0
3 years ago
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
3 years ago
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