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castortr0y [4]
2 years ago
10

An object moving with a speed of 5m/s comes to rest in 10s after the brakes are applied . What is the initial velocity​?

Physics
1 answer:
Flauer [41]2 years ago
8 0

Initial velocity is 5m/s.

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A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is
cricket20 [7]

Answer:

<em>The period of the motion will still be equal to T.</em>

<em></em>

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

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

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>

5 0
3 years ago
Opal adds 25 grams of salt to a one-liter glass beaker filled up to its volume mark with pure water. She stirs the water until t
Mars2501 [29]

By what i know i think that the answer would be A a homogeneous mixture.

5 0
3 years ago
A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from
krek1111 [17]

Answer:

E=35921.96N/C

Explanation:

From the question we are told that:

Radius r=0.321mm

Charge Density \mu=0.100

Distance d= 5.00 cm

Generally the equation for electric field is mathematically given by

E=\frac{mu}{2\pi E_0r}

E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}

E=35921.96N/C

4 0
3 years ago
A 2.50-m segment of wire carries 1000 A current and feels a 4.00-N repulsive force from a parallel wire 5.00 cm away. What is th
Stolb23 [73]

Answer:

The current is  I_b  =  400 \ A

Explanation:

From the question we are told that

    The  length of the segment is  l  =  2.50  \  m

     The current is  I_a  =  1000 \ A

     The force felt is  F  =  4.0 \  N

        The distance of the second wire is  d =  5.0 \ cm  = 0.05 \  m

Generally the current on the second wire is mathematically represented as

        I_b  =  \frac{2 \pi * r * F }{ l *  \mu_o  *  I_a }

Here  \mu_o is the permeability of free space with value  \mu_o =  4 \pi * 10^{-7} \ N/A^2

=>      I_b  =  \frac{2 * 3.142  *  0.05 *  4 }{ 2.50  *  4\pi *10^{-7}  * 1000 }

=>      I_b  =  400 \ A

4 0
3 years ago
There are three ways to give an object a static charge. They are friction, conduction and induction. Two pieces of cellophane ta
Novay_Z [31]

The tape is charged by friction and her hand has an induced opposite charge. They therefore attract.

8 0
3 years ago
Read 2 more answers
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