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Brilliant_brown [7]
2 years ago
10

Derive an expression for the acceleration of the car. Express your answer in terms of D and vt Determine the time at which the s

peed of the car is equal to the speed vt of the truck. Express your answer in terms of tD. Justify your answer.
Physics
1 answer:
Bezzdna [24]2 years ago
3 0

Answer:

V(car) = V(truck) at t = Dt/2

acceleration = v(car) = D/t^2

Explanation:

acceleration = v(car) = D/t^2

Since the average velocities must be the same, the car's final velocity must be twice the trunk velocity assuming the car start with zero velocity, since acceleration remain the same throughout the journey velocities at half-time point must be equal.

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A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = \frac{4}{3} \pi (\frac{D}{2})^3

Therefore \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

r of the disc = 1.15(\frac{ D}{2} )

Using conservation of angular momentum;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disc = m*\frac{   \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{  m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
2 years ago
Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a n
Savatey [412]

Answer:

The friend on moon will be richer.

Explanation:

We must calculate the mass of gold won by each person, to tell who is richer. For that purpose we will use the following formula:

W = mg

m = W/g

where,

m = mass of gold

W = weight of gold

g = acceleration due to gravity on that planet

<u>FOR FRIEND ON MOON</u>:

W = 1 N

g = 1.625 m/s²

Therefore,

m = (1 N)/(1.625 m/s²)

m(moon) = 0.6 kg

<u>FOR ME ON EARTH</u>:

W = 1 N

g = 9.8 m/s²

Therefore,

m = (1 N)/(9.8 m/s²)

m(earth) = 0.1 kg

Since, the mass of gold on moon is greater than the mass of moon on earth.

<u>Therefore, the friend on moon will be richer.</u>

7 0
2 years ago
A motorbike is traveling to the left with a speed of 27.0\,\dfrac{\text m}{\text s}27.0 s m ​ 27, point, 0, start fraction, star
myrzilka [38]

Answer:

\frac{729\frac{m^{2} }{s^{2} } }{83m}≈8.8\frac{m}{s^{2} }

Explanation:

7 0
2 years ago
A 5-ft-tall person walks away from the wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How
AveGali [126]

Answer:

The rate of change of the height is - 4 ft/s

Solution:

As per the question:

Height of the person, y = 5 ft

The rate at which the person walks away, \frac{dx}{dt} = 4\ ft/s

Distance of the spotlight from the wall, x = 40 ft

Now,

To calculate the rate of change in the height, \frac{dy}{dt} of the person when, x = 10 m:

From fig 1.

\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}

\frac{y}{40} = \frac{5}{x}

xy = 200                                                                       (1)

Differentiating the above eqn w.r.t time t:

x\frac{dy}{dt} + y\frac{dx}{dt} = 0

Thus

\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}              (2)

From eqn (1):

When x = 10 ft

10y = 200

y = 20 ft

Using eqn (2):

\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft

8 0
1 year ago
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