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3 years ago

How much current would flow in a device powered by four 1.5 V batteries through a bulb that had a resistance of 51.0 ohms?

Physics

1 answer:

Serhud [2]3 years ago

4 0

Answer:

0.12A

Explanation:

Total voltage produced by the batteries;

V = 1.5 × 4 = 6 v

Then the current that is flowing through the device

Current(I) = V/R

; I = 6/51

;Current = 0.12A

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What is the electric potential energy of a charge that experiences a force of 3.6x10^-4 N when it is 9.8x10^-5 m from the source

jekas [21]

Answer:

3.5 x 10^-8 J

Explanation:

We are given,

force experienced by a charge = 3.6 x 10^-4 N; and

distance of charge from the source of electric field = 9.8 x 10^-5 m

We know that, Energy = force x distance

therefore, to find the potential energy of the charge, we need to multiply its applied force with the given distance to get:

Electric potential energy of a charge = (3.6 x 10^-4) × (9.8 x 10^-5) = 3.5 x 10^-8 J

Therefore, the potential energy of a charge is 3.5 x 10^-8 J.

6 0

4 years ago

Please i need your help w/h/w you make the following measurements if an object:42kg, and 22m

Rufina [12.5K]

Answer:

$1.9 kg/m^3$

Explanation:

The density of an object is given by

$d=\frac{m}{V}$

where

m is the mass of the object

V is its volume

In this problem,

m = 42 kg

V = 22 m^3

Substituting into the equation, we find the object's density:

$d=\frac{42 kg}{22 m^3}=1.9 kg/m^3$

8 0

3 years ago

A carnival ride has a 2.0m radius and rotates once each .90s. Find the centripetal acceleration

Paraphin [41]

We know that centripetal acceleration is nothing but the ratio of the square of the tangential velocity to that of the radius vector.
a=v*v/r=ωωrr/r
=ωωr
=2πf2πfr
=2π2πr/TT
=97m/secsec

3 0

3 years ago

A person is standing on a level floor. His head, upper

BabaBlast [244]

Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity $= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}$

$y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}$

$y_{cg} = 1.03 m$

8 0

4 years ago

Which of the following is an ecosystem

iogann1982 [59]

A it is Green plants.

or they all

7 0

3 years ago