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3 years ago

The force of gravity on Mercury is less than the Earth because:

2 answers:

8 0

The answer is B.

Earth's mass is <span>5.972 × 10^24 kg
Mercury's mass is </span><span>3.285 × 10^23 kg

Hope i helped</span>

Vinvika [58]3 years ago

7 0

Mercury has less mass than earth. So the answer is B

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An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and

ladessa [460]

Answer:

The answer to your question is 5.5 x 10¹² kg/km³

Explanation:

Data

density = rho = X

mass = 5.94 x 10²⁴ kg

volume = 1.08 x 10¹² km³

Process

To solve this problem use the density formula and simplify it to find the result.

Formula

density = $\frac{mass}{volume}$

Substitution

density = $\frac{5.94 x 10^{24} }{1.08 x 10^{12}}$

Simplification and result

density = 5.5 x 10¹² kg/km³

5 0

3 years ago

A basketball has a mass of 601 g. Moving to the right and heading downward at an angle of 29 degree to the vertical, it hits the

olga55 [171]

Answer:

Change in momentum, $\Delta p=6.3\ kg-m/s$

Explanation:

It is given that,

Mass of the basketball, m = 601 g = 0.601 kg

The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s

It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :

$\Delta p=mv\ cos\theta-(-mv\ cos\theta)$

$\Delta p=2mv\ cos\theta$

$\Delta p=2\times 0.601\times 6\times \ cos(29)$

$\Delta p=6.3\ kg-m/s$

So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.

6 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.

irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0

3 years ago

V=I/R correctly expresses the relationship between voltage, current, and resistance.

docker41 [41]

The correct answer is true

5 0

1 year ago

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