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3 years ago

5

he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The

weight of a space probe on the surface of Mars is about _____.

1 answer:

Ahat [919]3 years ago

5 0

1/3 the weight than it is on earth

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How do you break apart a

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B

Explanation:

Depends Mostly on bonds electrolysis can be used, chemical bonding like additional of water or by heating back to their elements.

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2 years ago

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To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration

astra-53 [7]

In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.

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Two dogs fight over a bone. The larger of the two pulls on the bone to the right with a force of 42 N. The smaller one pulls to

tamaranim1 [39]

Answer:

Explanation:

A

35 N Small Dog <=======BONE=========> Bigger Dog 42 N

B

Fnet = Large Dog - small dog The forces are subtracted because they are acting in opposite Directions.

Fnet = 42 - 35

Fnet = 7 N

C

m = 2.5 kg

F = 7 N

a = ?

F = m * a

7 = 2.5 a

a = 7 / 2.5

a = 2.8 m/s^2

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2 years ago

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

storchak [24]

<span>On what:

f (is the focal length of the lens) = ?
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
$\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}$
$\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}$
$\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}$
$\frac{1}{f} = \frac{10}{33.18}$
Product of extremes equals product of means:
$10*f = 1*33.18$
$10f = 33.18$
$f = \frac{33.18}{10}$
$\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark$

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3 years ago

A 57 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.1 m/s. The acceleration of

kari74 [83]

Answer:

Her altitude as she crosses the bar, h₂ is approximately 6.1 m

Explanation:

The given parameters of the motion of the pole vaulter are;

The mass of the pole vaulter, m = 57 kg

The speed with which the pole vaulter is running, u = 11 m/s

The speed of the pole vaulter when she crosses the bar, v = 1.1 m/s

The acceleration due to gravity, g = 9.8 m/s²

From the total mechanical energy, M.E. equation, we have;

M.E. = P.E. + K.E.

Where;

P.E. = The potential energy of the motion = m·g·h

K.E. = The kinetic energy of the motion = 1/2·m·v²

By the principle of conservation of energy, we have;

The change (loss) in kinetic energy, ΔK.E. = The change (gain) in potential energy, ΔP.E.

ΔK.E. = 1/2·m·(v² - u²)

ΔP.E. = m·g·(h₂ - h₁)

Where;

h₁ = The ground level = 0 m

h₂ = The altitude with which she crosses the bar

∴ 1/2·m·(v² - u²) = m·g·(h₂ - h₁)

(h₂ - h₁) = (v² - u²)/(2·g) = (11² - 1.1²)/(2·9.8) = 6.11173469388

h₂ = 6.11173469388 + h₁ = 6.11173469388 + 0 = 6.11173469388

h₂ = 6.11173469388

Her altitude as she crosses over the bar, h₂ ≈ 6.1 m.

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3 years ago