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3 years ago

What is quantity?explain.

2 answers:

8 0

Answer: Quantity is the extent, size, or sum of an object or how much of that object there is.
For ex: A small quantity of milk or the oceans vast quantity of fish

WARRIOR [948]3 years ago

6 0

Answer:

the amount or number of a material or immaterial thing not usually estimated by spatial measurement

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Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$

6 0

3 years ago

A uniformly charged sphere has a total charge of 300uc and a radius of 8cm. Find the electric field density at A point 16cm from

s2008m [1.1K]

E = kQ 

(r + h)²

where,

k = 9 × 10^9Nm²C^-2

Q = total charge, 300uC = 300 × 10^ -6C

r = 8 × 10^ -2m

h = 16 × 10^ -2m

then,

E = 9e9 * 300e^-6 

(8e^-2 + 16e^-2)²

E = 4687500N/C

6 0

2 years ago

An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object

Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate = 50 -10 = 40 N

Mass of the object = 73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a a = 5.38 m/s^2

6 0

1 year ago

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1 2 and t = 1 s? Use G

Romashka-Z-Leto [24]

There is one mistake in the question.The Correct question is here

A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.

Answer:

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

Explanation:

Given data

time=1/2 sec to 1 sec

v(t)=-9.8t m/s

To find

Distance

Solution

As the acceleration as first derivative of velocity with respect to time

acceleration(-g)= dv/dt

Solve it

dv = a dt

dv = -g dt

v - v₀ = -gt

v= dy/dt

dy = v dt

dy = ( v₀ - gt ) dt

y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]

y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]

y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )

y(1s) - y(1/2s) = - 3.675 m

The cat falls 3.675 m between time 1/2 s and 1 s.

6 0

3 years ago

Can I PLEASE get some help? I REALLY need it!

soldi70 [24.7K]

The answer is C. Hope this helps.

7 0

3 years ago