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2 years ago

He chart below compares the characteristics of four objects, A, B, C, and D, discovered in the solar system.

Physics

1 answer:

Leto [7]2 years ago

5 0

Answer:

Object A and C (Second choice)

Let's observe one by one

#A..

Kupier belt is a belt associated at Neptune outerside similar to asteroid belt

Hence it's correct

#B.

Has enough gravity to keep other objects far away from its orbit

It's any planet or may be sun /star

Is in orbit around the sun

Sure it's correct

Is almost circular in shape

It's any planet or moons

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Which of the following is most likely to be an observation made by a physiologist

Naya [18.7K]

Do you have the answer choices ?

3 0

3 years ago

When you apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed ().When you apply a f

elena-s [515]

Answer:

Explanation:

When we apply a horizontal force of 76 N to a block, the block moves across the floor at a constant speed. So net force on the block is zero .

It implies that a force ( frictional ) acts on it which is equal to 76 N in opposite direction ( friction )

When we apply a greater force on it it starts moving with acceleration .

This time kinetic friction acts on it due to rough ground equal to 76 N .This is limiting friction ( maximum friction )

Net force on the body in later case

= 89 - 76

= 13 N

Force by ground on the block in horizontal direction = 76 N ( FRICTIONAL FORCE )

3 0

3 years ago

What is NOT a principle of genetics?

timurjin [86]

The recessive trait will always show up

3 0

3 years ago

Read 2 more answers

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

7 0

3 years ago

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

kramer

Answer:

$v_x = -6\ m/s$

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

$v_x = v cos \theta$

velocity is in negative x-direction, v = -12 m/s

now,

$v_x = -12\times cos 60^0$

$v_x = -12\times 0.5$

$v_x = -6\ m/s$

Hence, the velocity x-component is equal to -6 m/s.

5 0

3 years ago