A quantity y is to be determined from the equation y=(px)/q^2

Sketch the circuit labeling the meter and bulb as two separate resistors connected in parallel to the voltage source. Then show

Ksenya-84 [330]

Answer:

Show attached picture

Explanation:

Let's call V the voltage provided by the battery in the circuit. M is the multimeter (let's call $R_M$ its internal resistance) and R indicates the resistance of the light bulb.

We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

$R_M = 1000 R$ (1)

Both the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

$V_M = V_R$

Using Ohm's law, $V=RI$ , we can rewrite the previous equation as:

$R_M I_M = R I_R$

where

$I_M$ is the current in the meter

$I_R$ is the current in the bulb

Using (1), this equation becomes

$(1000 R) I_M = R I_R \rightarrow I_M = \frac{I_R}{1000}$

so, the current in the meter is 1000 times less than through the bulb.

5 0

1 year ago

Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en

trasher [3.6K]

If the maximum permissible limit for depression of the structure is 20 centimeters, the number of floors that can be safely added to the building is C. 18

depression = (depression/floor)(# floors) < 20

Here are the following choices:
A. 14
B. 15
C. 18
D. 23

8 0

1 year ago

(a) when rebuilding her car's engine, a physics major must exert 300 n of force to insert a dry steel piston into a steel cylind

Vilka [71]

There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03

Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
$F_f = \mu N$
where $\mu$ is the coefficient of friction, while N is the normal force. So we have:
$F=\mu N$
since we know that F=300 N and $\mu=0.3$ , we can find N, the magnitude of the normal force:
$N= \frac{F}{\mu}= \frac{300 N}{0.3}=1000 N$

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is $\mu=0.03$ due to the presence of the oil. Therefore, we have:
$N= \frac{F}{\mu}= \frac{300 N}{0.03}=10000 N$

8 0

1 year ago

A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that

gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken