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3 years ago

5

A circuit has a current of 1.2 A. If the voltage decreases to one-third of its original amount while the resistance remains cons

tant, what will be the reaulting current?

1 answer:

rosijanka [135]3 years ago

6 0

Answer:

1.2*1/3

Explanation:

Multiply the current by the volatage to get ur ans

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A refrigerator has a mass of 88 kg. What is the weight of the refrigerator?

posledela

Answer:

I don't think the information is complete

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2 years ago

A box mass of 24kg is being pulled horizontally on a rough surface by an applied force of 585N. The coefficient of kinetic frict

Elan Coil [88]

the normal force is the force applied opposite to the weight of the was box. So the normal force is equal to the weight of the box = 24 kg *(9.81 m/s2) = 235.44 N

the acceleration of the box be solve using newtons 2nd law of motion:

F = ma

a = F/ m = 585 N/ 24 kg = 24.38 m/s2

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3 years ago

Read 2 more answers

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago

Identify the subordinate clause and tell if it is used as a noun, adjective, or adverb. Click on the blue box until the correct

charle [14.2K]

The subordinate clause is "<span>who are loyal and industrious" and is used as an adjective to describe students.</span>

7 0

3 years ago

The total charge a battery can supply is rated in mA⋅h , the product of the current (in mA ) and the time (in h ) that the batte

natita [175]

Answer: 0.2 hours

Explanation: In order to solve this question we have to considerer that a recargeable battery can supply 1800 mA in one hour then we have to determine how long could this battery drive current through a long, thin wire of resistance 34 Ω .

Besides, this battery has a voltage of 12 V

so by using the Ohm law we also know that V=R*I,

Fron this we can obtain:

I= V/R= 12 V/ 34 Ω=0.35 A= 350 mA

then considering that this battery can supply 1800 mA in one hour we have this battery can supply 350 mA in x time in the form:

1hour------- 1800 mA

x hour--------350 mA

time= 350/1800= 0.2 hour

4 0

2 years ago