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Alchen [17]
3 years ago
14

16. Calculate the mass in kg of an object that is being accelerated 12 m/s with a force of 654 N.

Physics
2 answers:
Andrej [43]3 years ago
6 0

Mass=force/acceleration

Or,mass=654/12

Or,mass=54.5kg

So,the mass is 54.5kg

Cloud [144]3 years ago
3 0
(654m/s ) / (12m/s) = 54.5 kg.
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Please could someone explain this.
icang [17]

This problem is to let you practice using Newton's second law of motion:

Force = (mass) x (acceleration)

-- The airplane's mass when it takes off (before it burns any of its load of fuel) is 320,000 kg.

-- The force available is (240,000 N/per engine) x (4 engines) = 960,000 N.

-- Now you know ' F ' and ' mass '.  Use Newton's second law of motion to calculate the plane's acceleration.

7 0
3 years ago
billy bob joe's truck has a mass of 3,800 g and an acceleration of 4.5 m/s^2. what is the force of the truck? (you will have to
maria [59]

Answer:

<h2>17.1 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

3800 g = 3.8 kg

We have

force = 3.8 × 4.5

We have the final answer as

<h3>17.1 N</h3>

Hope this helps you

4 0
3 years ago
A lamp hangs from the ceiling at a height of 2.9 m. If the lamp breaks and falls to the floor, what is its impact speed
ser-zykov [4K]
To solve this there is this website that I found that helps
I am in middle school so I have no idea how to solve this
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this website may help considering u are in high school and u
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8 0
3 years ago
A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

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3 years ago
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3 years ago
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