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Yakvenalex [24]
4 years ago
9

A truck on a straight road starts from rest and accelerates at 1.4 m/s 2 until it reaches a speed of 22 m/s. Then the truck trav

els for 22 s at constant speed until the brakes are applied, stopping the truck in a uniform manner in an additional 4.4 s. How long is the truck in motion?
Physics
1 answer:
Zolol [24]4 years ago
3 0

Answer:

42.11 sec

Explanation:

In first part : The truck start from the rest that is u = 0 m/sec

Acceleration is given as a=1.4m/sec^2

It reaches to a velocity of 22 m/sec so v=22 m/sec

Now according to first law of motion v=u+at

22=0+1.4\times t

t=15.71sec

In second part : The truck maintain a constant speed for 22 sec

In third  part : The truck takes an additional 4.4 sec for stopping

So total time in which the truck is in motion =15.71+22+4.4=42.11 sec

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Which of the following terms relate to the meaning of "physical"
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6 0
3 years ago
Suppose you average 75 MPH over the first halfof a drive, and your average speed for the entiredrive is 60 MPH. What was your av
alexandr402 [8]

Answer:

x = 45 MPH

Explanation:

given,

Average speed of the first half = 75 MPH

Average speed of entire ride = 60 MPH

Average speed of the second half = ?

let the average speed of the second half = x MPH

now,

average of entire ride is given as 60 mph so,

 \dfrac{75+x}{2} = 60

 75+x= 2\times 60

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         x = 120 -75

        x = 45 MPH

hence, the average speed of the second half comes out to be 45 MPH.

5 0
4 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce
ollegr [7]
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
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v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
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7 0
3 years ago
An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line
Nata [24]

Answer:

10.22 cm

Explanation:

linear charge density, λ = 7.5 x 10^-12 C/m

distance from line, r = 14.5 cm = 0.145 m

initial speed, u = 3000 m/s

final speed, v = 0 m/s

charge on proton, q = 1.6 x 10^-19 C

mass of proton, m = 1.67 x 10^-27 kg

Let the closest distance of proton is r'.

The potential due t a line charge at a distance r' is given  by

V=-2K\lambda ln\left (\frac{r'}{r}  \right )

where, K = 9 x 10^9 Nm^2/C^2

W = q V

By use of work energy theorem

Work = change in kinetic energy

qV = 0.5m(u^{2}-v^{2})

By substituting the values, we get

V=\frac{mu^{2}}{2q}

-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}

- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }

- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }

- ln\left ( \frac{r'}{r} \right )=0.35

\frac{r'}{r} =e^{-0.35}

\frac{r'}{r} =0.7047

r' = 14.5 x 0.7047 = 10.22 cm

5 0
3 years ago
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