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Ghella [55]
3 years ago
13

Four toy racecars are racing along a circular race track. The cars start at the 3-o'clock position and travel CCW along the trac

k. Car A is constantly 2 feet from the center of the race track and travels at a constant speed. The angle Car A sweeps out increases at a constant rate of 1 radian per second.
Required:
How many radians θ does car A sweep out in t seconds?
Physics
1 answer:
xeze [42]3 years ago
6 0

Answer:

in t seconds, Car A sweep out t radian { i.e θ = t radian }

Explanation:

Given the data in the question;

4 toy racecars are racing along a circular race track.

They all start at 3 o'clock position and moved CCW

Car A is constantly 2 feet from the center of the race track and moves at a constant speed

so maximum distance from the center = 2 ft

The angle Car A sweeps out increases at a constant rate of 1 radian per second.

Rate of change of angle = dθ/dt = 1

Now,

since dθ/dt = 1

Hence θ = t + C

where C is the constant of integration

so at t = 0, θ = 0, the value of C will be 0.

Hence, θ = t radian

Therefore, in t seconds, Car A sweep out t radian { i.e θ = t radian }

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Sergio [31]

Answer:

Part a)

v_f = 25.2 m/s

t = 5.48 s

Part b)

v_f = 25.32 m/s

t = 4.96 s

Explanation:

Part a)

When ski start from rest

v_f^2 - v_i^2 = 2 a d

on this inclined plane we know that the acceleration is given as

a = g sin\theta

a = 9.81 sin28

a = 4.6 m/s^2

now for final speed

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(4.6)(69)

v_f = 25.2 m/s

now time taken by the ski to reach the bottom is given as

v_f = v_i + at

25.2 = 0 + 4.6 t

t = 5.48 s

Part b)

Now when ski start with initial speed of 2.5 m/s

then we will have

v_f^2 - v_i^2 = 2 a d

v_f^2 - 2.5^2 = 2(4.6)(69)

v_f = 25.32 m/s

now time taken by the ski to reach the bottom is given as

v_f = v_i + at

25.32 = 2.5 + 4.6 t

t = 4.96 s

3 0
3 years ago
In the southern hemisphere the summer solstice occurs when the sun is
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... directly over the Tropic of Capricorn, about 23.5 degrees south of the equator. The date is around December 22 or 23.
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An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
4 years ago
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AlexFokin [52]

Answer:

I just answered a question like this. It should be B. An electromagnet :)

Explanation:

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"<span>The particles of gases are the least capable of moving around"
its the opposite, gases have the most capability to move around
</span>
8 0
3 years ago
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