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2 years ago

If a circuit has a voltage of 250 volts and a resistance of 25 ohms, what are the amps for the current in the circuit?

Physics

2 answers:

Anna [14]2 years ago

7 0

Answer:

Given that,

Voltage (v) = 250 V ,

Resistance (R) = 25 Ω,

Determine current (I) =?

We know that from ohm's law

V = I R

I = V/R

= 250 / 25

I = 10 A

<em>Current in the circuit is 10 Amps</em>

gavmur [86]2 years ago

5 0

U=R•I=>I=U/R=250/25=10 A

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You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon

Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

$u_{1}$ = 0m/s

$t_{1}$ = t

$u_{2}$ = 43.12m/s

$t_{2}$ = (t-2.2)s

The distance by the first balloon is

$D = u_{1} t_{1} + \frac{1}{2} at_{1}^2$

where

a = 9.8m/s2

Inputting the values

$D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2$

The distance traveled by the second balloon

$D = u_{2} t_{2} + \frac{1}{2} at_{2}^2$

Inputting the values

$D = (43.12)(t-2.2) + \frac{1}{2} (9.8)(t-2.2)^2$

simplifying

$D = 4.9t^2 + 21.56t -71.148$

Substituting D of the first balloon into the D of the second balloon and solving

$4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s$

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

$D = 4.9(3.3)^2\\ D = 53.361 m$

7 0

2 years ago

If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v

dexar [7]

The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units

5 0

2 years ago

Which of the following statements are true about an object in two-dimensional projectile motion with no air resistance? (There c

ki77a [65]

Answer:

The correct answers are

The following statements are true about an object in two-dimensional projectile motion with no air resistance

D) The speed of the object is zero at its highest point.

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

Explanation:

A) The speed of the object is constant but its velocity is not constant.

False the vertical velocity increases on descent

B) The acceleration of the object is constant but its object is + g when the object is rising and -g when it is falling.

False, the acceleration is -g when the object is rising

C) The acceleration of the object is zero at its highest point.

False, the acceleration is constant in magnitude throughout the motion

D) The speed of the object is zero at its highest point.

True, the direction of motion changes at the highest point from hence the body comes to rest and the speed is zero

E) The horizontal acceleration is always zero and the vertical acceleration is always a non-zero constant downward

True, the horizontal acceleration has associated force during motion but the vertical acceleration is due to gravity which is constant downwards

6 0

2 years ago

Calculate the applied force of the washers on the car. First, convert the mass you recorded for one, two, three, and four washer

Andrej [43]

The mass of one washer is 0.0049 kg.

The mass of two washers is 0.0098 kg.

The mass of three washers is 0.0147 kg.

The mass of four washers is 0.0196 kg.

3 0

2 years ago

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A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

5 0

2 years ago

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