All categories

2 years ago

What is the current of a 10 Ohm resistor attached to a 9 volt battery?

1 answer:

7 0

I'm giving you a general formula , its called ohm law it states that electric current is proportional to voltage and inversely proportional to resistance

if V=voltage associated with the battery
R=resistance , I=current then
 $\boxed{V=I*R}$

You might be interested in

The energy emitted from the sun is a product of ________.

Tamiku [17]

Q. The energy emitted from the sun is a product of ________.

A. Fusion

3 0

3 years ago

which of the following professionals would most likely be called upon to treat a person suffering from a dangerous level of schi

Likurg_2 [28]

The primary physician in cases of schizophrenia is the psychiatrist.

hope this helps, Terrakmiller80! :-)

7 0

2 years ago

Consider the data collected in science class. Different masses were thrown with varied amounts of

lesya [120]

Answer:

A: In all cases, the acceleration was the same.

Explanation:

I know this because its a clear obvious answer not only that it was one of my USA TESTPREP questions and it was right.

All you mainly have to do is the math - F=ma , In each case , the acceleration is 5 m/s squared

4 0

2 years ago

Read 2 more answers

In the periodic table, similar elements are arranged in vertical columns called

tester [92]

Answer:

The vertical columns on the periodic table are called groups or families because of their similar chemical behavior.

Explanation:

3 0

2 years ago

Read 2 more answers

A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m

slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 $m/s^{2}$ and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

$M\frac{dv}{dt} = u\frac{dM}{dt}$

$\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt$

= $u\int_{M_{o}}^{M_{f}} \frac{dM}{M}$

$v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}$

$v_{o} = 0$

As, $v_{f} = u ln (\frac{M_{f}}{M_{o}})$

u = -2900 m/s

$M_{f} = M_{o} - m \times t_{f}$

= $2500 kg + 1000 kg - 95 kg \times t_{f}s$

= $(3500 - 95t_{f})s$

$v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s$

Also, we know that

a = $\frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}$

= $\frac{u}{3500 - 95 t} \times (-95) m/s^{2}$

= $\frac{95 \times 2900}{3500 - 95t} m/s^{2}$

At t = 10 sec,

$v_{f}$ = 918.34 m/s

and, a = 108 $m/s^{2}$

3 0

2 years ago