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2 years ago

A 263 kg motorcycle moves with a velocity of 5 m/s. What is its kinetic energy? ______ joules

1 answer:

7 0

$K.E= \frac{1}{2} mv^2$

K.E= Kinetic Engery
M=Mass
V=Velocity
m=263kg
v=5m/s

Plug the given info into the formula:
$K.E= \frac{1}{2} (263)(5)^2$

Final answer: <span>3287.5J </span>

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Someone plz help.The temperature started at 35°F. If the temperature went down 4°F and then up 2°F, whats is the temperature dif

timama [110]

Answer:

The temperature difference relative to the high temperature is -2°F.

Step-by-step explanation:

We are given that the temperature started at 35°F. The temperature went down 4°F and then up 2°F.

Firstly, the original temperature = 35°F

Now, it is stated that the temperature went down 4°F, this means that the temperature decreases by this amount.

So, the new temperature = 35°F - 4°F = 31°F

Now, the temperature went up by 2°F, this means that the temperature increases by this amount.

So, the final temperature = 31°F + 2°F = 33°F

Now, the final temperature difference relative to the high temperature is given by = 33°F - 35°F = -2°F.

{Here, the maximum temperature is 35°F}

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2 years ago

A circle has a sector with area 33 pi and a central angle of 11/6 pi radians. What is the area of a circle?

Mashutka [201]

Answer:

36π

Step-by-step explanation:

The area of a circle is given as:

$A = \pi r^2$

where r = radius of the circle

The area of a sector of a circle is given as:

$A_s = \frac{\alpha }{2\pi} * \pi r^2$

where α = central angle in radians

Since $\pi r^2$ is the area of a circle, A, this implies that:

$A_s = \frac{\alpha }{360} * A$

A circle has a sector with area 33 pi and a central angle of 11/6 pi radians.

Therefore, the area of the circle, A, is:

$33 \pi = \frac{\frac{11 \pi}{6} }{2 \pi} * A\\\\33\pi = \frac{11}{12} * A\\\\=> A = \frac{33\pi * 12}{11}\\ \\A = \frac{396 \pi}{11} \\\\A = 36\pi$

The area of the circle is 36π.

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Answer:

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Step-by-step explanation:

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