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3 years ago

(NEED HELP ASAP)

Physics

1 answer:

Mice21 [21]3 years ago

6 0

Answer:

a = 0.98J

Explanation:

PE = mgh

PE = 0.1 x 9.8 x 1

PE = 0.98J

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Think of a sphere with many a star-

ycow [4]

Answer:

For it is a Globular Cluster

Explanation:

These Globular clusters are not considered to be a galaxies because they bond gravitationally to orbiting galaxies like the Milky Way and their masses are relatively small.

In globular clusters, the stars are closed together and they are like 70times closer to one another in their solar neighbourhood.

A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers.

3 0

3 years ago

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.7 1010 m (inside the orbit

Lubov Fominskaja [6]

Answer:

58515.9 m/s

Explanation:

We are given that

$d_1=4.7\times 10^{10} m$

$v_i=9.5\times 10^4 m/s$

$d_2=6\times 10^{12} m$

We have to find the speed (vf).

Work done by surrounding particles=W=0 Therefore, initial energy is equal to final energy.

$K_i+U_i=K_f+U_f$

$\frac{1}{2}mv^2_i-\frac{GmM}{d_1}=\frac{1}{2}mv^2_f-\frac{GmM}{d_2}$

$\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2}=\frac{1}{2}v^2_f$

$v^2_f=2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})$

$v_f=\sqrt{2(\frac{1}{2}v^2_i-\frac{GM}{d_1}+\frac{GM}{d_2})}$

Using the formula

$v_f=\sqrt{v^2_i+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

$v_f=\sqrt{(9.5\times 10^4)^2+2\times 6.7\times 10^{-11}\times 1.98\times 10^{30}(\frac{1}{6\times 10^{12}}-\frac{1}{4.7\times 10^{10})}$

Where mass of sun= $M=1.98\times 10^{30} kg$

$G=6.7\times 10^{-11}$

$v_f=58515.9 m/s$

4 0

3 years ago

According to Boyle’s Law, increasing the pressure of a gas will have which effect on its volume? increase decrease random effect

umka21 [38]

Volume will decrease if the heat remains constant

3 0

4 years ago

How far (in meters) above the earth's surface will the acceleration of gravity be 85.0 % of what it is on the surface?

suter [353]

Answer:

X = 6910319.7 m

Explanation:

let X be the distance where the acceleration of gravity is 85% of what it is on the surface and g1 be the acceleration of gravity at the surface and g2 be the acceleration of gravity at some distance X above the surface.

on the surface of the earth, the gravitational acceleration is given by:

g1 = GM/(r^2) = [(6.67408×10^-11)(5.972×10^24)]/[(6371×10^3)^2] = 9.82 m/s^2

at X meters above the earth's surface, g2 = 85/100(9.82) = 8.35m/s^2

then:

g2 = GM/(X^2)

X^2 = GM/g2

X = \sqrt{GM/g2}

= \sqrt{(6.67408×10^-11)(5.972×10^24)/ 8.35

= 6910319.7 m

Therefore, the acceleration of gravity becomes 85% of what it is on the surface at 6910319.7 m .

4 0

3 years ago

Two carts with masses of 4.2 kg and 3.2 kg move toward each other on a frictionless track with speeds of 5.4 m/s and 4.5 m/s, re

rodikova [14]

Answer:the final speed is 5.01 m/s

Explanation:

Momentum is the product of mass and velocity.

Cart 1 has a mass of 4.2kg and a speed 5.4 m/s

Cart 2 has a mass of 3.2kg and a speed 4.5 m/s

Total momentum before collision is

m1u1 + m2u2. It becomes

4.2×5.4 + 3.2×4.5 = 22.68 + 14.4

= 37.08kgm/s

The carts stick together after colliding head-on. This means that they move with a common velocity, v. Therefore, Total momentum after collision is (m1 + m2)v. It becomes

(4.2 + 3.2)v = 7.4v

According the the law of conservation of momentum, the total momentum before collision = the total momentum after collision. Therefore,

7.4v = 37.08

v = 37.08/7.4 = 5.01 m/s

8 0

3 years ago