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Anuta_ua [19.1K]
3 years ago
8

1 point

Physics
1 answer:
AysviL [449]3 years ago
4 0
I think it’s 150 because if u multiply 3 x 50 it’s 150 but so it’s ore simple you do 3 x5 equals 15 then add the zero from the 50 so you’ll get 150 hope I helped
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Who were we in the space/arms race with?<br> In the movie *Hidden figures*
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Explanation:

The U.S. launched its first man into space in May 1961.

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3 years ago
A ball thrown by Ginger is moving upward through the air. Diagram A shows a box with a downward arrow. Diagram B shows a box wit
vichka [17]

As the ball is moving in air as well as we have to neglect the friction force on it

So we can say that ball is having only one force on it that is gravitational force

So the force on the ball must have to be represented by gravitational force and that must be vertically downwards

So the correct FBD will contain only one force and that force must be vertically downwards

So here correct answer must be

<em>Diagram A shows a box with a downward arrow. </em>

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3 years ago
A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc
Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
2 years ago
A low-pass first-order instrument has a time constant of 20 ms. find the frequency, in hertz, of the input at which the output w
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?????????????????????????????
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How many signifficant figures do the following number have 258​
Romashka [77]

Answer:

<u>Facts about 258</u>

<u>Sig Figs</u>

3

<u>258</u>

<u>Decimals</u>

0

<u>Scientific Notation</u>

2.58 × 102

<u>E-Notation</u>

2.58e+2

<u>Words</u>

two hundred fifty-eight

Explanation:

258 Rounded to Fewer Sig Figs

2 260                2.6 × 102

1 300           3 × 102

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2 years ago
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