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sergey [27]
3 years ago
11

you are trying to figure out the voltage of a battery when you know the current and the resistance. the current(i) is 25 Amps, a

nd the resistance (R) is 10 ohms. Use the right form of I=VIR
Physics
2 answers:
Brums [2.3K]3 years ago
8 0

Answer:

V= 250 volts

Explanation:

V=IR

V=25*10

V=250 volts

dalvyx [7]3 years ago
6 0

Answer: 250 V

Explanation: Ohm's Law is expressed in the given equation:

V = I x R

Substitute the values of I ( current) and R ( resistance)

V = 25 A x 10 ohms

= 250 V

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True or false
ElenaW [278]
If im not mistaken it is true
4 0
3 years ago
A cycle travels along a circular track of diameter 42 m. Calculate the distance travelled and the displacement of the cycle in (
DENIUS [597]

Answer:

(a) i) The distance travelled by the cycle in half round is approximately 65.97 m

ii) The displacement is 42 m

(b) (i) The distance travelled in one round is approximately 131.95 m

(ii) The displacement of the cycle in one round is 0

Explanation:

The diameter of the track through which the cycle travels, D = 42 m

(a) i) Half round is the motion of half the length of the circular path

The distance travelled by the cycle in half round = The length of half the circular track = (1/2) × π × D

∴ The distance travelled by the cycle in half round = (1/2) × π × 42 m = 21·π m ≈ 65.97 m

ii) The displacement half round = The change in the location of the cycle = The difference between the start and stop locations of the cycle on a straight line after half round

The angle at the center of the circular path the cycle turns in half round  = 180°

Therefore, the path between the start and stop location of the cycle in half round = The diameter of the circular track

The displacement of the cycle in half round = The diameter of the circular track = 42 m

(b) (i) The distance travelled in one round = The perimeter of the circular track = π·D

∴ The distance travelled in one round = π × 42 m ≈ 131.95 m

(ii) The displacement of the cycle in one round = The change in the location of the cycle

The start and stop location of the cycle after moving one round is the same, therefore, there is no change in the location of the cycle.

Therefore we have;

The displacement of the cycle in one round = 0 (no change in location of the cycle)

7 0
3 years ago
In the expression 3x2y + 4x – 5x2y, are the terms 3x2y and –5x2y like terms? Explain by providing support for your answer.
Zanzabum
Https://www.cymath.com/answer?q=%205%2B4i%2F6%2B8i
this site can help you 
3 0
3 years ago
Read 2 more answers
A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder
Stella [2.4K]

Answer:

The temperature change per compression stroke is 32.48°.

Explanation:

Given that,

Angular frequency = 150 rpm

Stroke = 2.00 mol

Initial temperature = 390 K

Supplied power = -7.9 kW

Rate of heat = -1.1 kW

We need to calculate the time for compressor

Using formula of compression

\terxt{time for compression}=\text{time for half revolution}

\terxt{time for compression}=\dfrac{1}{2}\times T

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{f}

Put the value into the formula

\terxt{time for compression}=\dfrac{1}{2}\times \dfrac{1}{150}\times60

\terxt{time for compression}=0.2\ sec

We need to calculate the rate of internal energy

Using first law of thermodynamics

U=Q-W

\dfrac{\Delta U}{\Delta t}=\dfrac{\Delta Q}{\Delta t}-\dfrac{\Delta W}{\Delta t}

Put the value into the formula

\dfrac{\Delta U}{\Delta t}=(-1.1)-(7.9)

\dfrac{\Delta U}{\Delta t}=6.8\ kW

We need to calculate the temperature change per compression stroke

Using formula of rate of internal energy

\dfrac{\Delta U}{\Delta t}=\dfrac{nc_{v}\Delta \theta}{\Delta t}

\Delta\theta=\dfrac{\Delta U}{\Delta t}\times\dfrac{\Delta t}{n\times c_{c}}

Put the value into the formula

\Delta \theta=6.8\times10^{3}\dfrac{0.2}{2.0\times20.93}

\Delta\theta=32.48^{\circ}

Hence, The temperature change per compression stroke is 32.48°.

6 0
4 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
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