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2 years ago

Under what condition(vector A + vector B)= (vector A) + (vector B) holds good?

Physics

1 answer:

romanna [79]2 years ago

6 0

Answer:

when angle between them is zero degree

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speed of sound is 343 Ms at 20 degrees Celsius. The frequency heard from the sound is 256 Hz. what is the sounds wavelength?

Lina20 [59]

S= 343m/s
F=256Hz

WL= 343ms/256-1
WL=V/F

= 1.339844m

7 0

2 years ago

A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from point A to point B along

Voltage difference,ΔV =V₁ - V₂

The work done

W = Q . ΔV

Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.

ΔV = 0

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0

2 years ago

Can someone please help me?

svetlana [45]

Answer:

acceleration...............

7 0

3 years ago

Read 2 more answers

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago

calculate the amount of heat (in btu) needed to raise the temp of 2.5lb of glass from 45°f to 350°f. the specific heat capacity

Firlakuza [10]

So you would use the equation Q=cmΔT, where c is the specific heat, m is the mass, and ΔT is change in temperature. Q, or heat added, would equal (0.187)(2.5)(350-45), which simplifies to 142.5875 btu.

7 0

2 years ago