If

A battery has a potential resistance of 12 V and a current of 1280 mA. What is the resistance? PLS HELP ME ASAPPPPPPPPPP

Romashka-Z-Leto [24]

Resistance (R) = <u>9.375 ohm (Ω)</u>

Steps:

$R = \frac{V}{I}$

$= \frac{12 \: volt}{1280 \: milliampere}$

$= \frac{12 \: volt}{1.28 \: ampare}$

$= 9.375 \: ohm (Ω)$

6 0

2 years ago

As an interstellar cloud of hydrogen gas shrinks in size, its rate of rotation

FinnZ [79.3K]

Answer:

INCREASES, BECAUSE ITS ANGULAR MOMENTUM IS CONSERVED.

Explanation: Interstellar cloud of Hydrogen is an accumulation of Hydrogen gas in the cloud.

As the Interstellar cloud of Hydrogen shrinks (reduces) in size,the rate of rotation of the shrinked Interstellar cloud Increases because its angular momentum is conserved. GASEOUS MOLECULES MAKE UP ABOUT 99% OF THE INTERSTELLAR CLOUD WITH HYDROGEN HAVING ABOUT 90% OF THE VOLUME OF GASES IN THE INTERSTELLAR CLOUD.

4 0

2 years ago

When you must give something up in order to get something else, it is called...

natali 33 [55]

Opportunity cost refers to what you have to give up to buy what you want in terms of other goods or services. When economists use the word “cost,” we usually mean opportunity cost.

6 0

2 years ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago

A 16.2 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

S_A_V [24]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

$\sum F_x = 0$

$f-N_w = 0$

$N_w = f$

The forces in the horizontal direction would be,

$\sum F_y = 0$

$N_f -W =0$

$N_f = W$

The sum of Torques at equilibrium,

$\sum \tau = 0$

$Wdcos\theta - N_wLsin\theta = 0$

$WdCos\theta = fLSin\theta$

$f = \frac{Wd}{Ltan\theta}$

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

$f_{max} = \mu W=\frac{Wd}{Ltan\theta}$

$\theta = tan^{-1} (\frac{d}{\mu L})$

Replacing,

$\theta = tan^{-1} (\frac{0.9}{0.42*2})$

$\theta = 46.975\°$

Therefore the minimum angle that the person can reach is 46.9°

8 0

2 years ago