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weqwewe [10]
3 years ago
9

Please...physics rotation questions

Physics
1 answer:
kari74 [83]3 years ago
7 0
1st Question:
There are 4 formulas involving angular velocity:
For the first question, we are given the initial angular velocity (ωo), final angular velocity (ωf), and the amount of time. The question is asking us to find the angular acceleration (α).
Knowing this information, we can use w=wo+ \alpha t
Lets plug in the values given in the question:
w=wo+ \alpha t
123  \frac{rad}{sec}  = 11  \frac{rad}{sec} + (4.5 sec) \alpha
Now we just solve for \alpha
123 \frac{rad}{sec} = 11 \frac{rad}{sec} + (4.5 sec) \alpha
Subtract 11 from each side
112 \frac{rad}{sec} = (4.5 sec) \alpha
Divide each side by 4.5 sec. 
\alpha = 24.89  \frac{rad}{sec^{2} }
----------------------------------------------------------
2nd Question:
For the second question, we want to find the change in distance (which is AKA radians). So we know we would need to use either the 1st, 3rd or the 4th equation. We are given time (t=45 s), the initial velocity (wo=0 rad/s), and the final velocity (w=245 rad/s). Since the 4th equation doesn't have time, and the 3rd equation has angular acceleration (which we weren't given), we know we will need to use the 1st equation. 
theta=wt (in which w is the change in velocity. Which for this case, would be 245-0)
theta=(245 rad/s)(45 s) 
Simplify
theta=11025 rad 

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The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r
Goshia [24]

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength \lambda =1.4cm =0.014m

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m

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4 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
3 years ago
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