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3 years ago

What is meant by a quantity ? Mention any three system of unit that you know to measure a

Physics

1 answer:

d1i1m1o1n [39]3 years ago

4 0

 

 the quantity which can be measured is called 

 the quantity which can be measured is called quantity for example length and mass time temperature 

 the quantity which can be measured is called quantity for example length and mass time temperature please brainlist 

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I think I know this one but I don’t at the same time

galben [10]

Answer: B.) 6

Explanation:

To answer this problem you get the number of students who attended Wednesday (18) and the number of students who attended Tuesday (12) and subtract 18 - 12 = 6

Answer = B.) 6

8 0

3 years ago

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

3 years ago

Using the data table to the right, determine the

andreyandreev [35.5K]

Silver sable is in spiderman lol

6 0

3 years ago

Read 2 more answers

A uniform electric field has a magnitude 1.80 kV/m and points in the +x direction. (a) What is the electric potential difference

EastWind [94]

Answer:

a)ΔV = 6.48 KV

b)ΔU =18.79 mJ

Explanation:

Given that

E= 1.8 KV/m

We know that

Electric potential difference ΔV given as

ΔV = E .d

Here

E= 1.8 KV/m

d= 3.6 m

ΔV = E .d

ΔV = 1.8 x 3.6 KV

ΔV = 6.48 KV

Given that

q=+2.90 µC

Change in electric potential energy ΔU given as

ΔU = q .ΔV

$\Delta U=2.9\times 10^{-6}\times 6.48\times 10^3\ J$

ΔU =18.79 mJ

8 0

3 years ago

If the car goes exits a freeway and goes from 65 mph to 35 mph is it accelerating?

Zolol [24]

Answer:

No, the car is decelerating

Explanation:

No the car is decelerating if it exits a freeway and goes from 65

mph to 35 mph since the change in velocity is negative.

change in velocity = final - initial

change in velocity = 35 - 65

change in velocity = -30mph

Since the change in velocity is negative, hence the car is decelerating. Deceleration is a negative acceleration

8 0

3 years ago