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3 years ago

Which of the following is the same size as Pluto?

Physics

2 answers:

Cloud [144]3 years ago

5 0

The Moon I believe correct me if I'm wrong.

coldgirl [10]3 years ago

4 0

I would say it's <span>C. the Moon
</span>

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Your GPS shows that your friend’s house is 10.0 km away. But there is a big hill between your houses and you don’t want to bike

kati45 [8]

Answer:

The value is $c = 8 \ km$

Explanation:

From the question we are told that

The distance of friends house from your point is $a = 10 \ km$

The distance of your friends street from your street is $b = 6 \ km \ in the \ direction \ towards \ the \ north$

The diagram illustrating this question is shown on the first uploaded image

From the diagram we can apply by Pythagoras theorem as follows

$a^2 = b^2 + c^2$

=> $c = \sqrt{^2 - b^2}$

=> $c = \sqrt{ 10^2 - 6^2}$

=> $c = 8 \ km$

6 0

4 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

My name is Ann [436]

Answer:

915m

Hope this helps.

5 0

3 years ago

A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the roc

vovangra [49]

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

$v_x = 2.70 m/s$

Part b)

In vertical direction we can use kinematics

$v_f = v_i + at$

$v_f = 0 + 2.1 * 9.8$

$v_f = 20.6 m/s$

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

5 0

3 years ago

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t

ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0

3 years ago

Consider a double slit experiment in the air. The wavelength of light is 500nm light, the screen is 1m away and two adjacent bri

Bess [88]

Answer: $0.75\ cm$

Explanation:

Given

Wavelength of light $\lambda=500\ nm$

Screen is $D=1\ m$ away

Distance between two adjacent bright fringe is $\Delta y=\dfrac{\lambda D}{d}$

When same experiment done in water, wavelength reduce to $\dfrac{\lambda }{\mu}$

So, the distance between the two adjacent bright fringe is $\Delta y'=\dfrac{\lambda D}{\mu d}$

Keeping other factor same, distance becomes

$\Rightarrow \dfrac{1}{\frac{4}{3}}=\dfrac{3}{4}\quad \text{Refractive index of water is }\dfrac{4}{3}\\\\\Rightarrow 0.75\ cm$

3 0

3 years ago