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2 years ago

A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *

Physics

1 answer:

Oksi-84 [34.3K]2 years ago

7 0

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

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Determine the torque applied to the shaft of a car that transmits 225 hp

Arisa [49]

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

$T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }$

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

$T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu$

8 0

2 years ago

What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2

kkurt [141]

Answer:

$4.3\cdot 10^{-12} J$

Explanation:

The fusion reaction in this problem is

$4^1_1H \rightarrow ^4_2He +2e^+$

The total energy released in the fusion reaction is given by

$\Delta E = c^2 \Delta m$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

$m(^1_1H)=1.007825u$ is the mass of one nucleus of hydrogen

$m(^4_2 He)=4.002603u$ is the mass of one nucleus of helium

So the mass defect is:

$\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u$

The conversion factor between atomic mass units and kilograms is

$1u=1.66054\cdot 10^{-27}kg$

So the mass defect is

$\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg$

And so, the energy released is:

$\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J$

6 0

2 years ago

If period of the pendulum in preceding sample problem were 24s how tall would the tower be ?

frutty [35]

Answer:

So length of pendulum is 143.129 m

Explanation:

We have given period of simple pendulum is 2 sec

We have to find the length of simple pendulum

Let the length of pendulum is l

Acceleration due to gravity $g=9.8m/sec^2$ is

Time period is given by $T=2\pi \sqrt{\frac{l}{g}}$

So $24=2\times 3.14\times \sqrt{\frac{l}{9.8}}$

$\sqrt{\frac{l}{9.8}}=3.821$

Squaring both side

${\frac{l}{9.8}}=14.60$

l =143.129 m

So length of pendulum is 143.129 m

8 0

2 years ago

Read 2 more answers

If you wanted to measure the width of the gym, how would the accuracy of a meter stick compare with that of a 50m tape

Alisiya [41]

Meter stick would not be as accurate,
Every time you placed it down and picked it back up you run the chance of losing 2-4 cm each time.

7 0

2 years ago

A firecracker is thrown downward from a height of 2.75m above the ground, with a speed of 3.15m/s. Ignore air resistance, determ

3241004551 [841]

Here in this question as we can see there is no air friction so we can use the principle of energy conservation

$PE_i + KE_i = PE_f + KE_f$

$mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2$

now here we know that

$h_1 = 2.75 m$

$v_i = 0$

$v_f = 5.23 m/s$

now plug in all values in above equation

$mg*2.75 + 0 = mgh + \frac{1}{2}m(5.23)^2$

divide whole equation by mass "m"

$9.8*2.75 = 9.8*h + \frac{1}{2}*27.35$

$9.8*h = 13.27$

$h = 1.35 m$

so height of the ball from ground will be 1.35 m

4 0

2 years ago