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3 years ago

What’s the potential difference across a 5.0 ohms resistor that carries a current of 5.0A

Physics

1 answer:

vivado [14]3 years ago

3 0

Answer:

The correct answer is B-25 V

Explanation:

We apply Ohm's Law, according to which:

V = i x R

V = 5A x 5Ω

V= 25 V

Being V the potential difference whose unit is the VOLT, i the current intensity (Ampere) and R the electrical resistance (ohm)

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A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monke

sasho [114]

Answer:

Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.

If the cannon throws the banana with the same force the monkey falls

(m.g=Fz <=> m.9,81N/kg=...N).

Then the throw will slow down because of the gravitational pull.

Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.

The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.

If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.

So to do so you need to throw the bananas with a speed of at least 9,81m.s

Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.

I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.

kind regards

4 0

3 years ago

a 0.199 kg snowball moving west makes an inelastic collision with a 2.89 kg box moving 0.523 m/s west. afterward,they move west

kogti [31]

Answer:

The initial velocity of the snowball was 22.21 m/s

Explanation:

Since the collision is inelastic, only momentum is conserved. And since the snowball and the box move together after the collision, they have the same final velocity.

Let $m_1$ be the mass of the ball, and $v_1$ be its initial velocity; let $m_2$ be the mass of the box, and $v_2$ be its velocity; let $v_f$ be the final velocity after the collision, then according to the law of conservation of momentum:

$m_1v_1+m_2v_2=v_f(m_1+m_2)$ .

From this we solve for $v_1$ , the initial velocity of the snowball:

$\boxed{v_1=\frac{v_f(m_1+m_2)-m_2v_2}{m_1}}$

now we plug in the numerical values $m_1=0.199\:kg$ , $m_2=2.89\:kg$ , $v_2=0.523\:m/s$ , and $v_f=1.92\:m/s$ to get:

$v_1=\frac{1.92*(0.199+2.89)-2.89*0.523}{0.199}$

$\boxed{v_1=22.21\:m/s}$

The initial velocity of the snowball is 22.21 m/s.

P.S: we did not take vectors into account because everything is moving in one direction—towards the west.

4 0

3 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

3 years ago

Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o

o-na [289]

Answer:

20.08 Volts

Explanation:

Parallel Connection of Capacitors

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

$\displaystyle V_1=\frac{Q_1}{C_1}$

$\displaystyle V_2=\frac{Q_2}{C_2}$

They are both the same after connecting them, thus

$\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}$

Or, equivalently

$\displaystyle Q_2=\frac{C_2Q_1}{C_1}$

The total charge of both capacitors is

$\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)$

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

$Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C$

Now we compute Q1 from the equation above

$\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C$

The final voltage of any of the capacitors is

$\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V$

7 0

3 years ago

What effect does the sun have on Earth's tides? A. The pull of the sun cancels the pull of the moon when they are on opposite si

Kisachek [45]

Answer:

Explanation:

because two vectors which align in the same line adds one to another

6 0

3 years ago