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Serga [27]
3 years ago
8

According to ohms law, as the voltage increases across a 40 ohm resistor what happens to the current, resistors, and resistance

Physics
1 answer:
irakobra [83]3 years ago
7 0

Answer:

As the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

Explanation:

Ohm's law states that the current flowing through a circuit is directly proportional to the applied voltage.

V = IR

where;

I is the current

R is the resistance

V is the applied voltage

Based on this law (Ohm's law), as the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

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What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

8 0
3 years ago
An 800 N man stands on a scale in a motionless elevator. When the elevator begins to move, the scale
DENIUS [597]

Answer:

Net force = 800 – 650

= 150 N

150 = (800 ÷ 9.8) a  

a = 1470 ÷ 800

= 1.8375 m/s^2, downwards

3 0
3 years ago
a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this
Yuri [45]

Answer:

-32.5 * 10^-5 J

Explanation:

The potential energy of this system of charges is;

Ue = kq1q2/r

Where;

k is the Coulumb's constant

q1 and q2 are the magnitudes of the charges

r is the distance of separation between the charges

Substituting values;

Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)

Ue= -32.5 * 10^-5 J

4 0
3 years ago
Which law represents a balanced chemical equation?
neonofarm [45]
Law of conservation of mass
6 0
3 years ago
Read 2 more answers
The chart show the masses and velocities of two colliding objects that stick together after a collision.
finlep [7]

Answer:

<u><em></em></u>

  • <u><em>1,500 kg.m/s</em></u>

Explanation:

First, arrange the information in a table:

Object        Mass (kg)           Velocity (m/s)

  A                    200                      15

  B                     150                    - 10

After the collision, the two objects are stick together, thus you talk aobut one object and one momentum.

According to the law of convervation of momentum, the momentum after the collision is equal to the momentum before the collision.

<u>Momentum before the collision, P₁</u>:

          P_1=m_{A,1}\times v_{A,1}+m_{B,1}\times v_{B,1}

         P_1=200kg\times 15m/s+150kg\times (-10m/s)\\\\  P_1=3000kg.m/s-1500kg.m/s=1500kg.m/s

<u>Momentum after the collision</u>:

  • As stated, it es equal to the momentum before the collision: 1,500 kg . m/s
6 0
3 years ago
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