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3 years ago

Superman is flying 54.5 m/s when he sees

Physics

1 answer:

olchik [2.2K]3 years ago

6 0

Answer:

Acceleration is 69.6 m/s²

Explanation:

From the second newton's equation of motion:

${ \boxed{ \bf{s = ut + \frac{1}{2} a {t}^{2} }}}$

substitute the variables:

${ \tt{850 = (54.5 \times 4.22) + ( \frac{1}{2} \times a \times {4.22}^{2}) }} \\ { \tt{850 = 229.99 + 8.9042a}} \\ { \tt{8.9042a = 620.01}} \\ { \tt{a = 69.6 \: {ms}^{ - 2} }}$

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Satellite C revolves around Earth 10 times a day. What is the radius of its orbit, measured from Earth's center? Assume that Ear

frez [133]

The radius of its orbit, measured from Earth's center, will be 1.44 × 10⁷ mm.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data in problem is;

The mass of Earth is, $\rm m_E = 5.98 \times 10^{24} \ kg$

Gravitational constant, G =6.674 × 10⁻¹¹ N m₂/kg²

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

$\rm F_g = \frac{Gm_sm_e}{r^2}$

The centripetal force due to rotation of the satellite;

$\rm F_c = \frac{m_s v^2}{r}$

The centripetal and the gravitational force are equal;

$\rm F_g = F_c \\\\ \frac{Gm_sm_e}{r^2} = \frac{m_s v^2}{r} \\\\ r = G \frac{m_E }{v^2 } \\\\ r = G \frac{m_E }{(r \omega )^2 } \\\\ r = \sqrt[3]{\frac{Gm_E}{\omega^2}} \\\\ r = \sqrt[3]{\frac{6.67 \times 10^{-11}(5.98 \times 10^{24})}{(3.63 \times 10^{-4})^2}} \\\\ r = 1.44 \times 10^7 \ mm$

Hence, the radius of its orbit measured from Earth's center will be 1.44 × 10⁷ mm.

To learn more about Newton's law of gravitation, refer to the link.

brainly.com/question/9699135.

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8 0

2 years ago

A machine has an efficiency of 80%. How much work must be done on the machine so to make it do 50,000 J of output work?

Tamiku [17]

Work formula is Work=N∙M or Joule (J)

So you have the following given:

50 000 is the output work

.8 is the efficiency

if you input the given = 50000 = .8*J

To get the answer just divide 50000 by .8 to get the answer.

62,500 J is the amount of work to be done.

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3 years ago

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Hydrogen fluoride gas (HF) and sodium hydroxide (NaOH) react in a test tube. They form water and sodium fluoride (NaF). Which ty

AfilCa [17]

This reaction is a double replacement reaction. Hydrogen in the hydrogen fluoride gets replaced by sodium of the sodium hydroxide. HF is a weak acid and NaOH is a strong base. So, double replacement produces sodium fluoride and water molecules.

6 0

3 years ago

The gravity of Earth is attracting a person towards the center with 500N of gravitational force. The person is exerting a reacti

mestny [16]

Answer:

500n

Explanation:

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3 years ago

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Problem 1: Three beads are placed along a thin rod. The first bead, of mass m1 = 24 g, is placed a distance d1 = 1.1 cm from the

Svet_ta [14]

Answer:

b) $x_{cm}$ = 4.88 cm , c) $x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃) and d)

$x_{cm}$ ’= 1.88 cm

Explanation:

The definition of mass center is

$x_{cm}$ = 1/M ∑ xi mi

Where mi, xi are the mass and distance from an origin for each mass and M is the total mass of the object.

Part b

Apply this equation to our case.

Body 1

They give us the mass (m₁ = 24 g) and the distance (d₁ = 1.1 cm) from the origin at the far left

Body 2

They give us the mass (m₂ = 12.g) and the distance relative to the distance of the body 1, let's look for the distance from the left end (origin)

D₂ = d₁ + d₂

D₂ = 1.1 + 1.9

D₂ = 3.0 cm

Body 3

Give the mass (m₃ = 56 g) and the position relative to body 2, let's find the distance relative to the origin

D₃ = D₂ + d₂

D₃ = 3.0 + 3.9

D₃ = 6.9 cm

With this data we substitute and calculate the center of mass

M = m₁ + m₂ + m₃

M = 24 + 12 + 56

M = 92 g

$x_{cm}$ = 1/92 (1.1 24 + 3.0 12 + 6.9 56)

$x_{cm}$ = 1/92 (448.8)

$x_{cm}$ = 4,878 cm

$x_{cm}$ = 4.88 cm

This distance is from the left end of the bar

Par c)

In this case we are asked for the same calculation, but the reference system is in the center marble, we have to rewrite the distance with the reference system in this marble.

Body 1

It is at d1 = -1.9 cm

It is negative for being on the left and the value is the relative distance of 1 to 2

Body 2

d2 = 0 cm

The reference system for her

Body 3

d3 = 3.9 cm

Positive because that is to the left of the reference system and is the relative distance between 2 and 3

Let's write the new center of mass (xcm')

$x_{cm}$ ’= 1/M (m₁ d₁ + m₂ d₂ + m₃ d₃)

$x_{cm}$ ’= 1/M (m₁ d₁ + m₃ d₃)

Part d) Let's calculate the value of the center of mass

$x_{cm}$ ’= 1/92 ((24 (-1.9) +56 3.9)

$x_{cm}$ ’= 1/92 (172.8)

$x_{cm}$ ’= 1.88 cm

This distance is to the right of the central marble

3 0

3 years ago