All categories

3 years ago

Help asap!

Physics

1 answer:

patriot [66]3 years ago

7 0

Answer:

220 V

Explanation:

$r_{1} = 20 ohm\\r_{2} = 30ohm\\r_{3} = 60ohm$

As the resistors are connected in series,

$R_{eq} = r_{1} +r_{2} +r_{3}$

= 20 + 30 + 60

= 110 ohms

V = IR

= 2 * 110

= 220 V

Hope it helps

You might be interested in

A tank is full of oil weighing 40 lb/ft^3. The tank is an inverted right circular cone (with the base at the top) with a height

krok68 [10]

Answer:

26945.6 ft⋅lbf

Explanation:

Volume of Right Circular Cone = pi*(radius^2)*(height/3)

Pi*(4)*(5/3) = 20.94 ft^3

Density = Mass / Volume

Mass = Density*Volume

Mass = (40)*(20.94)

Mass = 837.6 lb

Work = Force*Height

Force = Mass*Acceleration

Acceleration will be gravitational acceleration

Work = (837.6)*(32.17)*(1)

Work = 26945.6 ft⋅lbf

8 0

3 years ago

PLEASE HELP!! I WILL MARK BRAINLIST

Stella [2.4K]

sound wave amplitude

3 0

3 years ago

A basketball has a mass of 567 g. Heading straight downward, in the direction, it hits the floor with a speed of 2 m/s and rebou

Vsevolod [243]

Answer:

$\Delta p=2.27\frac{kg\cdot m}{s}$

Explanation:

The momentum change is defined as:

$\Delta p=p_f-p_i\\\Delta p=mv_f-mv_i\\\Delta p=m(v_f-v_i)(1)$

Taking the downward motion as negative and the upward motion as positive, we have:

$v_f=2\frac{m}{s}(2)\\v_i=-2\frac{m}{s}(3)$

Replacing (2) and (3) in (1):

$\Delta p=567*10^{-3}kg(2\frac{m}{s}-(-2\frac{m}{s}))\\\Delta p=2.27\frac{kg\cdot m}{s}$

5 0

3 years ago

Model rocket A is launched at an angle of 70° above the x axis, with an initial velocity of 40 m/s. An identical rocket B is lau

amid [387]

<span>A. Rocket A will travel farther horizontally than rocket B.

This is because from the x axis, 40 m/s at 90 degrees travels directly vertical. 40 m/s at 70 degrees is slightly horizontal, so it will travel further horizontally.</span>

3 0

3 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$