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wlad13 [49]
3 years ago
6

A 50 kg wagon is pulled with a 100 N force. If the force is increased to 200 N, how quickly will the wagon accelerate? 1 m/s2 2

m/s2 4 m/s2 25 m/s2
Physics
2 answers:
motikmotik3 years ago
6 0

Answer:

4 m per second

Explanation:

Katen [24]3 years ago
3 0
4 m/s2

I hope this helps!
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An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.
sergejj [24]

Answer:\sqrt{2}T

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

T=2\pi \sqrt{\frac{m}{k}}

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

T'=2\pi \sqrt{\frac{2m}{k}}

T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}

T'=\sqrt{2}T

i.e. New time period becomes \sqrt{2} times of previous one

                         

7 0
2 years ago
A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate
kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid R=1.06 m

Emf induced E=8.50\times 10^{-6} V/m

no of turns per meter n=450

we know Induced EMF is given by

\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A

Magnetic Field is given by

B=\mu _0ni

thus \frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}

Area of cross-section

A=\pi R^2 where

solving integration we get

E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2

where r=distance from axis

R=radius of Solenoid

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}

\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s

4 0
3 years ago
Found throughout the nervous system, __________ aid and support the function of neurons.
Nonamiya [84]

Answer:

Glia

Explanation:

8 0
2 years ago
Read 2 more answers
Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Zielflug [23.3K]

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

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3 years ago
Over thousands of years, when sediment is squeezed together and the pore space between the grains is reduced, this is the proces
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This is the process of compaction.
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