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Liula [17]
3 years ago
5

A remote-controlled car’s wheel accelerates at 22.2 rad/s2 . If the wheel begins with an angular speed of 11.0 rad/s, what is th

e wheel’s angular speed after exactly twelve full turns? Answer in units of rad/s.
Physics
1 answer:
wariber [46]3 years ago
4 0

The angular speed remained constant after twelve turns at 11 rad/s.

Explanation:

As per the Newtons equation of motion, the final velocity can be determined by using the below equation, provided ,the displacement and acceleration with initial velocity is given.

v^{2}= u^{2}+2as

Since, the present case follows circular motion, the displacement will be equal to angular displacement. And the velocity is equal to angular velocity.

As the car revolves complete twelve turns, then the displacement will be zero, since the final position is equal to the initial position.

So,

v^{2}= (11)^{2}+2*22.2*0\\\\v = 11 rad/s

Thus, the angular speed remained constant after twelve turns at 11 rad/s.

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Two car collide in an intersection. The speed limit in that zone is 30 mph. The car (mass of 1250 kg) was going 17.4 m/s (38.9).
Musya8 [376]

Answer:

u₂ = 3.7 m/s

Explanation:

Here, we use the law of conservation of momentum, as follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\

where,

m₁ = mass of the car = 1250 kg

m₂ = mass of the truck = 2020 kg

u₁ = initial speed of the car before collision = 17.4 m/s

u₂ = initial speed of the tuck before collision = ?

v₁ = final speed of the car after collision = 6.7 m/s

v₂ = final speed of the truck after collision = 10.3 m/s

Therefore,

(1250\ kg)(17.4\ m/s)+(2020\ kg)(u_2)=(1250\ kg)(6.7\ m/s)+(2020\ kg)(10.3\ m/s)\\\\(2020\ kg)(u_2) = 8375\ N.s + 20806\ N.s - 21750\ N.s\\\\u_2=\frac{7431\ N.s}{2020\ kg}

<u>u₂ = 3.7 m/s</u>

5 0
3 years ago
The sound waves from a noisy jet travel from the air into water. Which property of the wave will not change?​
notka56 [123]

The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.

Hope this helps! :)

8 0
3 years ago
Rank these temperatures from hottest to coldest: 32° F,32° C, and 32 K 320 F&gt; 32° C&gt;32 K 32°C 32° F 32 K 32° F 32 K 32° c
Leviafan [203]

Answer:

32 C > 32 F > 32 K

Explanation:

32 F, 32 C, 32 K

Let T1 = 32 F

T2 = 32 C

T3 = 32 K

Convert all the temperatures in degree C

The relation between F and C is given by

(F - 32) / 9 = C / 100

so, (32 - 32) / 9 = C / 100

C = 0

So, T1 = 32 F = 0 C

The relation between c and K is given by

C = K - 273 = 32 - 273 = - 241

So, T3 = 32 K = - 241 C

So, T 1 = 0 C, T2 = 32 c, T3 = - 241 C

Thus, T2 > T1 > T3

32C > 32 F > 32 K

3 0
3 years ago
A man 2 meters tall walks at the rate of 2 meters per second toward a streetlight that's 5 meters above the ground. At what rate
GaryK [48]

Answer:

It changes at a rate of 4/3 meter per second

Explanation:

In the given figure below we have

\Delta OBD\simeq \Delta ABC\\\\\therefore \frac{5}{X+Y}=\frac{2}{Y}\\\\

Solving for Y given  X=2m/s we get

\frac{5}{2+Y}=\frac{2}{Y}\\\\5Y=4+2Y\\\\Y=\frac{4}{3}m/s

8 0
3 years ago
A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe
sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let \alphabe the angular acceleration

2 T\times r=I\times \alpha

2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha

\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}

\alpha =\frac{3.24\times 10^3}{m} rad/s^2

\omega =\omega _0+\alpha \cdot t

\omega =0+\frac{3.24\times 10^3}{m}\times 0.74

\omega =\frac{2.4\times 10^3}{m} rad/s

Angular momentum

L=I\omega

L=0.5\times mr^2\times \omega

L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}

L=0.01051 kg-m^2/s

4 0
3 years ago
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