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3 years ago

The asteroid belt is located between Mars and Jupiter and contains thousands of asteroids.

Physics

2 answers:

Virty [35]3 years ago

6 0

Im not understanding the question but this is true

Wewaii [24]3 years ago

6 0

This is true but what's the question?

You might be interested in

A car initially at rest accelerates at 10m/s^2. The car’s speed after it has traveled 25 meters is most nearly... A.) 0.0m/s B.)

STALIN [3.7K]

The car traverses a distance $x$ after time $t$ according to

$x=\dfrac12at^2$

where $a$ is its acceleration, 10 m/s^2. The time it takes for the car to travel 25 m is

$25\,\mathrm m=\left(5\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=\sqrt 5\,\mathrm s$

5 is pretty close to 4, so we can approximate the square root of 5 by 2. Then the car's velocity $v$ after 2 s of travel is given by

$v=\left(10\dfrac{\rm m}{\mathrm s^2}\right)(2\,\mathrm s)\approx20\dfrac{\rm m}{\rm s}$

which makes C the most likely answer.

3 0

4 years ago

A plane flies 1800 miles in 9 hours, with a tailwind all the way. the return trip on the same route, now with a headwind, tak

Fittoniya [83]

Initially its moving with tail wind so here the speed of wind will support the motion of the plane

so we can say

$V_{plane} + v_{wind} = \frac{distance}{time}$

$V_{plane} + v_{wind} = \frac{1800}{9}$

$V_{plane} + v_{wind} = 200 mph$

now when its moving with head wind we can say that wind is opposite to the motion of the plane

$V_{plane} - v_{wind} = \frac{distance}{time}$

$V_{plane} - v_{wind} = \frac{1800}{12}$

$V_{plane} - v_{wind} = 150mph$

now by using above two equations we can find speed of palne as well as speed of wind

$V_{plane} = 175 mph$

$v_{wind} = 25 mph$

5 0

3 years ago

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

$\dfrac{1}{v}=\dfrac{187}{3741}$

$v=20.0\ cm$

Hence, The image distance is 20.0 cm.

5 0

3 years ago

When work is done and a force is transferred which choice describes the movements of the object?

Arisa [49]

When work is done and a force is transferred an object must move

3 0

4 years ago

A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20

Klio2033 [76]

Answer:

$\boxed{{\boxed{\blue{ 12.5}}}}$

Explanation:

Given, for girl : Weight or force;

$\rm \: F_1 = 50 \: kgf$

Area of both heels;

$\rm \: A_1 = \; 2 ×1 \; cm^2 = 2 \: cm^2$

$\rm \: Pressure \: P_1 = \cfrac{F_1}{ A_1 } = \dfrac{50 \: kgf}{2 \: cm {}^{2} } = 25 \: kgf \: cm {}^{ - 1}$

For elephant, Weight = Force $\rm F_2$ = 2000 kg•f

Area of 4 feet;

$\rm \: A_2 = \; 4 \times 250 \; cm^2 = 1000 \: cm^2$

$\rm \: Pressure \: P_2 = {F_2}/{A_2} \; = \cfrac{2 \cancel{0 00 }\: kgf}{1 \cancel{000} \: cm^2} = 2 \: kgf \: { \:cm}^{- 1}$

Now;

$\rm = \dfrac{Pressure \: Exerted \: by \: the \: Girl}{Pressure \: exerted \: by \: the \: elephant}$

$= \rm \: P_1/P_2$

$\implies \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } = \rm\cfrac{25 \: \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}$

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0

2 years ago