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2 years ago

A well chosen lifetime activity is something that should hold a person's interest for a long time.

Physics

1 answer:

Masteriza [31]2 years ago

4 0

Answer:

true

Explanation:

as long as you are interested, you are happy

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A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of

lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = $20.6\times 10^3\ m$

$\theta$ = The angle subtended by moon = $9.6\times 10^{-3}\ rad$

Distance traveled is given by

$s=r\times\theta$

$\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}$

$\Rightarrow s=197.76\ m$

The distance traveled by the jet is 197.76 m

8 0

2 years ago

There are stars located in the center bulge of the Milky Way and the spiral arms of the Milky Way. What is the difference betwee

nadya68 [22]

Answer:

The stars at the center bulge are bigger and brighter than the stars in the arms.

Explanation:

5 0

3 years ago

What do you like about science?

Inessa [10]

Answer:

nngh have

bjruh hjrhhj be rnrnnrnrnnnrnjrjnnnnnnnnrnrn n n and I was nrn

Explanation:

jbbbbbhhhjjnnnnnnnnnnjhvcc

Explanation:

5 0

3 years ago

A motorbike reaches a speed of 20 m/s over 60m, whilst

Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

An equation of uniformly accelerated motion

V = vo + at

Vt² = vo² + 2a (x-xo)

x = distance on t

vo / vi = initial speed

vt / vf = speed on t / final speed

a = acceleration

vf=20 m/s

d = 60 m

a = 3 m/s²

$\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s$

7 0

3 years ago

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

lubasha [3.4K]

Answer:

$\rho = 12580.7 kg/m^3$

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

$F = \frac{GMm}{(r + h)^2}$

here we have

$F =\frac {mv^2}{(r+ h)}$

$\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}$

here we have

$v = \sqrt{\frac{GM}{(r + h)}}$

now we can find time period as

$T = \frac{2\pi (r + h)}{v}$

$T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}$

$1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}$

$M = 4.54 \times 10^{23} kg$

Now the density is given as

$\rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}$

$\rho = 12580.7 kg/m^3$

8 0

2 years ago