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2 years ago

Calculate the current flowing if a charge of 36 kilocoulombs flows in 1 hour.

Physics

2 answers:

dusya [7]2 years ago

5 0

Answer:

Current = 10 A

Explanation:

here's the solution :-

=》

$current \: = \frac{charge}{time}$

=》

$\frac{36000}{3600}$

=》

$10$

so, the current flowing = 10 Ampere

Crazy boy [7]2 years ago

4 0

Answer:

sorry I didn't get it, my apologies

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The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

4 0

3 years ago

Read 2 more answers

A mass of 8 kg moves at a rate of 14.3 m/sec. what is the KE developed by the mass?

Lynna [10]

Kinetic energy is the energy associated with the motion of an object. It's a scalar quantity, there is no direction associated with KE and it has no components. $KE = \frac{mv ^{2} }{2} = \frac{8 *14.3 ^{2} }{2} = 4 *204.49 = 817.96J$ . Therefore Kinetic energy is 817.96J.

6 0

3 years ago

The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote

Alenkasestr [34]

It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..

8 0

3 years ago

An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror

andrew-mc [135]

Answer:

Position = $\frac{60}{7}\ cm$ behind the mirror

Nature = Virtual and Erect

Size = $\frac{15}{7}\ cm$ : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm$

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = $=\frac{h_{image}}{h_{object}}=$ $-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}$

Height of the object = 5 cm

Height of the image = $5\times\frac{3}{7}=\frac{15}{7}\ cm$

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0

3 years ago

An eagle flew for 3 hours at 115 km/h and 5 hours at 136 km/h. How far did the eagle fly?

Yanka [14]

Answer:

the eagle flew 1,025 km

Explanation:

since the eagle flew for 3 hours at 115 km/h it flew a total of 345 km during that time. During the 5 hours at 136 km/h the eagle flew a total of 680 km.

3 0

3 years ago