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Kryger [21]
3 years ago
8

A body travel from rest and acceleration to a speed of 120kg/hours in 10 seconds . calculate acceleration​

Physics
1 answer:
Vesna [10]3 years ago
6 0

Answer:

3.34kg/min²

Explanation:

Convert to the same unit of time:

120kg/h = 2kg/min

10s = 0.6min

acceleration = change in velocity / time

a = 2/0.6 = 3.34

Change unit if needed!

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The gravity of Neptune is about 1.1 times the gravity of earth. How will the mass of an object on Neptune compare with its mass
Roman55 [17]
I think the gravity doesn't affect the mass of an object. Only it's weight can be compared
7 0
3 years ago
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How much total energy is dissipated in 10. seconds
noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

P=\dfrac{E}{t}

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, P=I^2R

So equation (1) becomes :

E=I^2Rt

E=(0.5\ A)^2\times 4\Omega \times 10\ s

E=10\ J

So, the correct option is (3)

Hence, this is the required solution.

7 0
3 years ago
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You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H
Karo-lina-s [1.5K]

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end x= 3m, frequency of the wave f = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ v = fλ

Where v = speed of the standing wave.

also, ∴ v=\frac{x}{t}

where t = time take to fill entire length of the string.

Compare above both equation,

⇒   t = \frac{3}{2.3} sec

     t = 1.3sec

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

7 0
3 years ago
How do I find the approximate velocity of "the object", on the graph at 5 seconds?
MakcuM [25]

First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.

Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.


7 0
3 years ago
The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t
balu736 [363]

Answer:

D=2.996\times 10^{-2} m

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as  1.0\times 10^5 N/C.  transferring 3.9\times 10^9 electrons, the disk's Area can be calculated using the formula:

E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2

#We now calculate the disks diameter:

A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m

Hence, the diameter of the disks is D=2.996\times 10^{-2} m

8 0
3 years ago
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