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Rus_ich [418]
3 years ago
6

Help pls, see picture. Will mark Brainliest

Physics
1 answer:
Leno4ka [110]3 years ago
4 0

Answer:

B

Explanation:

the graph shows the line going up (accelerating) and it isn't curving like d so it doesn't stop accelerating

Hope this helps :)

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How many nanoseconds are there in 1.90 yr ?<br> Express your answer using three significant figures.
shepuryov [24]

       (1.9 yr) x (365.24 day/yr) x (86,400 sec/day) x (10⁹ nsec/sec)

   =  (1.9 x 365.24 x 86,400 x 10⁹) nanosec

   =  6.00 x 10¹⁶ nanoseconds

5 0
3 years ago
Mickey had three substances. He hit each of them with a hammer. A red
Verizon [17]

Answer:

A & D

Explanation:

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6 0
3 years ago
A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t
borishaifa [10]

Answer:

Angular acceleration will be 18.84rad/sec^2

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity \omega _i=0rev/sec

And final angular velocity \omega _f=24rev/sec

Time is given as t = 8 sec

From equation of motion

We know that \omega _f=\omega _i+\alpha t

24=0+\alpha \times 8

\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2

So angular acceleration will be 18.84rad/sec^2

4 0
3 years ago
Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
A race car makes one lap around a track of radius 50 m in 9.0 s. What is the average velocity? *
Oksi-84 [34.3K]

Given that,

Radius of track, r = 50 m

time , t = 9 s

velocity, v = ?

Distance covered by car in one lap around a track is equal to the circumference of the track.

C = 2 π r = 2 * 3.14 * 50

C = 314.159 m

Distance covered by car, s = 314.159 m

Velocity = distance/ time

V = 314.159 / 9

V = 34.9 m/s

The average velocity of car is 34.9 m/s.

7 0
3 years ago
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