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zhenek [66]
2 years ago
6

What is the mass of a bicycle that has 90 kg.m/s of momentum at a speed of 6 m/s?

Physics
1 answer:
OLEGan [10]2 years ago
6 0

Answer:

m = 15 kg

Explanation:

p = m × v

90 = m × 6

6m = 90

m = 90/6

m = 15 kg

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How much work must be done by frictional forces in slowing a 1000-kg car from 26.1 m/s to rest? a.3.41 x 10^5 J b.2.73 x 10^5 J
algol13

Answer:

Work done by the frictional force is 3.41\times 10^5\ J

Explanation:

It is given that,

Mass of the car, m = 1000 kg

Initial velocity of car, u = 26.1 m/s

Finally, it comes to rest, v = 0

We have to find the work done by the frictional forces. Work done is equal to the change in kinetic energy as per work - energy theorem i.e.

W=k_f-k_i

W=\dfrac{1}{2}m(v^2-u^2)

W=\dfrac{1}{2}\times 1000\ kg(0^2-(26.1\ m/s)^2)

W = −340605 J

or

W=3.41\times 10^5\ J

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6 0
3 years ago
Check the correctness of formula t=2π √m/k, dimensionally​
pantera1 [17]

Hi there! Lets see!

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  • k is the elastic constant measured in newtons per meter (N/m), or kilograms per second squared kg/s²

Therefore:

\sqrt{\dfrac{m}{k}} =\sqrt{\dfrac{[kg]}{[\dfrac{kg}{s^2}]}}  =\sqrt{\dfrac{[kg]}{[kg]}\cdot s^2} = \sqrt{[s]^2} = s

The period is given in seconds so the formula is dimensionally​ correct.

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3 years ago
Why are the oldest parts of the ocean floor no older than about 200 million years
ivann1987 [24]
The reason is because the molecules didnt come togeather around that time 

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A proton traveling along the x-axis enters a region at x = 0 where the x-component of the electric field is given by E = Ao/x1/2
storchak [24]

.Answer:

The value of the work done is \bf{ 5.29 qA_{0}}.

Explanation:

When a charged particle having charge q is moving through an electric field E, the net force (F) on the charge is

F = qE~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

and the work done (W) by the particle is

W = \int\limits^x_0 {F} \, dx ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Given, E = \dfrac{A_{0}}{x^{1/2}}.

Substitute the value of electric field in equation (1) and then substitute the result in equation (2).

W &=& \int\limits^7_0 {q\dfrac{A_{0}}{x^{1/2}}} \, dx \\&=& qA_{0} \int\limits^7_0 {x^{-1/2}} \, dx \\&=& 2qA_{0}[x^{1/2}]_{0}^{7}\\&=& 5.29 qA_{0}

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What cell processes occour during interphase
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