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3 years ago

How much voltage is required to run 0.42 A of current through a 150

Physics

2 answers:

IRINA_888 [86]3 years ago

8 0

Answer:

like the other person said, the answer is 6.3

Explanation:

igomit [66]3 years ago

4 0

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

$R=150 \Omega$ is the resistance

Substituting into the equation, we find the voltage needed:

$V=(0.42)(150)=63 V$

Learn more about voltage and current:

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A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. Whe

Allisa [31]

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

8 0

3 years ago

An ion with charge of Q = +3.2 x 10-19 C is in a region where a uniform electric field of magnitude E = 5.0 X 105 V/m is perpend

san4es73 [151]

Answer:

option C

Explanation:

given,

Q = +3.2 x 10⁻¹⁹ C

E = 5.0 X 10⁵ V/m

B = 0.80 T

ion's acceleration is zero

when acceleration is zero the magnitude of both the forces becomes equal.

q E = q V B

v = $\dfrac{E}{B}$

v= $\dfrac{5 \times 10^5}{0.80}$

v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s

hence, the correct answer is option C

8 0

3 years ago

PLEASE HELP ASAP!!!

musickatia [10]

Answer:

22.2 W

Explanation:

First of all, we calculate the work done by moving the wagon, using the formula:

$W=Fdcos \theta$

where

F = 20 N is the magnitude of the force

d = 1000 m is the displacement of the wagon

$\theta=0^{\circ}$ is the angle between the direction of the force and of the displacement (assuming the force is applied in the direction of motion)

Substituting, we find

$W=(20)(1000)=20,000 J$

Now we can find the power generated, which is equal to the ratio between the work done and the time taken:

$P=\frac{W}{t}$

where

W = 20,000 J

t = 15 min = 900 s

Substituting,

$P=\frac{20000}{900}=22.2 W$

And the same value in Joules/second (remember that 1 Watt = 1 Joule/second)

5 0

3 years ago

Match the following vocabulary words with their definitions:

Ronch [10]

Correct matching:

1 acceleration --> rate of change in velocity, which is the change in velocity divided by the change in time

2. speed --> the rate at which an object changes position when traveling in a certain direction

4. gravity --> force of attraction between all masses in the universe

5. Inertia --> an object´s resistance to a change in motion

3. friction --> force of resistance acting between objects in contact and tending to dampen their motion

6. velocity --> the rate at which an object changes position

6 0

3 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

(a). We need to calculate the work done on the object

Using formula of work done

$W=F\cdot s$

Put the value into the formula

$W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)$

$W=-13.6+55.8-0.53$

$W=41.67\ J$

(b). We need to calculate the average power due to the force during that interval

Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

W = work

t = time

Put the value into the formula

$P=\dfrac{41.67}{5.40}$

$P=7.71\ Watt$

(c). We need to calculate the angle between vectors

Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c). The angle between vectors is 79.7°

7 0

3 years ago