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3 years ago

How would the absence of gravity affect the formation of planets?

Physics

2 answers:

Doss [256]3 years ago

5 0

Answer : The rocks and debris would never accrete into a planet.

Explanation : Gravity is an attraction force and dependents on the mass of the objects but in the absence of the force the rocks and debris would not come together to form a clump.

So, we can say that in the absence of gravity, the rock and debris would never accrete into a planet.

Maru [420]3 years ago

3 0

The correct answer is B. The rocks and debris would never accrete into a planet.

Explanation:

Gravity refers to a force of attraction between objects due to their masses. This force is described through the formula F= G M m/ r2 in which G is equal to the gravitational constant, M and m describe the mass of objects, and r refers to the distance.

Moreover, gravity was essential for the formation of planets, stars and other elements in the universe, because this caused gas, dust, rocks, and other particles or elements to aggregate or accrete into planets and other celestial bodies. Moreover, gravity is responsible for phenomena such as planets orbiting. The previous idea implies that without gravity the formation of planets would not have possible and therefore "the rocks and debris would never accrete into a planet".

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A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly s

BARSIC [14]

Answer:

The axial force is $P = 15.93 k N$

Explanation:

From the question we are told that

The diameter of the shaft steel is $d = 50mm$

The length of the cylindrical bushing $L =100mm$

The outer diameter of the cylindrical bushing is $D = 70 \ mm$

The diametral interference is $\delta _d = 0.005 mm$

The coefficient of friction is $\mu = 0.2$

The Young modulus of steel is $207 *10^{3} MPa (N/mm^2)$

The diametral interference is mathematically represented as

$\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}$

Where $P_B$ is the pressure (stress) on the two object held together

So making $P_B$ the subject

$P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}$

Substituting values

$P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }$

$P_B = 5.069 MPa$

Now he axial force required is

$P = \mu * P_B * A$

Where A is the area which is mathematically evaluated as

$\pi d l$

So $P = \mu P_B \pi d l$

Substituting values

$P = 0.2 * 5.069 * 3.142 * 50 * 100$

$P = 15.93 k N$

8 0

3 years ago

3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) movin

Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0

2 years ago

Is hydrogen energy renwable? Why<br> How is it used nowadays<br> How is it obtained?

hram777 [196]

Answer:

Explanation:

5 0

2 years ago

DEVELOP SCIENCE CONCEPTS What kind of habitat do animals in<br> your community live in?

hammer [34]

Answer:

We live in a very adaptable environment. If we cannot adapt to the environment, we will not be able to survive.

Explanation:

5 0

2 years ago

Một dây dẫn đặt trong không khí có dòng điện I = 12A chạy qua, được gấp thành hình vuông cạnh a = 10cm. Xác định vectơ cường độ

disa [49]

Answer:

The net magnetic field ta the center of square is $1.36\times10^{-4} T$ .

Explanation:

Current, I = 12 A , side ,a = 10 cm = 0.1 m

Let the magnetic field due to the one side is B.

The magnetic field is given by

$B = \frac{\mu o}{4\pi}\times \frac{I}{r}\times \left (Sin A +Sin B \right )\\\\B = 10^{-7}\times \frac{12}{0.05}\times \left ( sin 45 + sin 45 \right )\\\\B = 3.4\times 10^{-5} T$

Net magnetic field at the center of the square is

B' = 4 B

$B'= 4\times 3.4\times 10^{-5}\\\\B' = 1.36\times10^{-4} T$

4 0

2 years ago