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iogann1982 [59]
3 years ago
11

A car takes 0.766 hours to drive 72.0 km to Georgia. How fast is the car going? Please help

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
Pretty fast. Everything looks fast when running past a light pole
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AlekseyPX

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48m, hope this helps :)

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4 years ago
Amy put some salt in a weighing dish. Then she put the weighing dish on a sensitive balance to measure the mass of the salt. She
tatyana61 [14]

Answer:

Explanation:

i. Accuracy in measurement is a term that is used to describe the closeness of a measured value to its actual value. It relates the actual value to the measured values.

Therefore in the given question, Amy's measurement of the salt's mass is accurate because her final value is close to all the measured values.

ii. Precision in measurement is a term that is used to describe the closeness of different measured values to each other. It relates the keenness in the measured values.

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3 years ago
How are energy losses reduced
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8 0
4 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
Please help
adelina 88 [10]

Answer:

I'm pretty sure that the answer is A

5 0
3 years ago
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