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Jobisdone [24]
2 years ago
14

Can an ordinary object, like a motorcycle, be mass-less? Yes or No

Physics
1 answer:
Drupady [299]2 years ago
4 0

Answer:

no.

Explanation:

because the mass of an object never changes.

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A car is at velocity of 20 km/h. If the car traveled 120 km in 3 hours at constant acceleration, what is its final velocity?
dedylja [7]
The velocity is dueuueue. 100
8 0
3 years ago
A pair of toy freight cars, one twice the mass of the other, fly apart when a compressed spring that joins them is released. Acc
saul85 [17]

Answer:

greater acceleration is experienced by the car with lower mass

Explanation:

Since both the toys are connected by same spring so the force due to spring on both the toys will be same and it is given as

F = kx

now we know by Newton's II law

F = ma

so here we have

a = \frac{F}{m}

here we have same force on both the blocks

so acceleration will be more if mass is less

so greater acceleration is experienced by the car with lower mass

8 0
3 years ago
A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
   185N = (mass) x (9.81 m/s²)
   Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u> 

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³.  So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

           Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

           Volume of displaced water = <u>3,058 cm³</u>

Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.


3 0
2 years ago
An object dropped from rest from the top of a tall building on planet x falls a distance d(t)18 left parenthesis t right parenth
melamori03 [73]

displacement is given by equation

d = 18t^2

now at t = 5 s the position is

d_1 = 18 *5^2 = 450 m

similarly position at t = 9 s

d_2 = 18*9^2 = 1458 m

so the displacement of object in given interval of time will be

d = 1458 - 450 = 1008 m

time interval

\delta t = 9 - 5 = 4 s

now the average velocity will be given as

v = \frac{\delta x}{\delta t}

v = \frac{1008}{4} = 252 m/s

so its average speed is 252 m/s

4 0
2 years ago
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
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