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2 years ago

The resistance of a lamp that draws 0.5A of current when it is operated by a 12V battery will be Ω

Physics

2 answers:

Vedmedyk [2.9K]2 years ago

7 0

According to ohm's law
V=IR
Or R=V/I
R=12/0.5
R=24ohm

lisabon 2012 [21]2 years ago

5 0

According to Ohm's Law, the resistance, current and voltage are related as:

V = IR
⇒
R = V/I

V is given to be 12 Volts.
I is given to be 0.5 Ampere

So, resistance will be:

R = 12/0.5 = 24 ohms

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A dogsled team is shown pulling a man on a sled. Below the picture is a free body diagram with 4 force vectors. The first vector

Scorpion4ik [409]

Answer:

yes

Explanation:

its on edg.

8 0

2 years ago

Read 2 more answers

A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown

nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

$v_x = 15.0 m/s$

and with an initial vertical velocity of

$v_{y0} = -20.0 m/s$

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

$y(t) = h + v_{0y} t + \frac{1}{2}gt^2$ (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

$0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m$

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

$v_y = -20 m/s$

So the vertical position of the balloon after a time t is

$y(t) = h + v_y t$

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

$y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m$

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

$v_x = 15.0 m/s$

So the horizontal distance travelled in t = 6.00 s is

$d_x = v_x t = (15.0 m/s)(6.00 s)=90 m$

Considering also that the vertical height of the balloon after t=6.00 s is

$d_y = 176.6 m$

The distance between the balloon and the rock can be found by using Pythagorean theorem:

$d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m$

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer at rest in the basket is

$v_y (t) = gt$

Substituting time t=6.00 s, we find

$v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s$

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

$v_x = 15.0 m/s$

Instead, the vertical velocity of the rock for an observer on the ground is now given by

$v_y (t) = v_{0y} + gt$

Substituting time t=6.00 s, we find

$v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s$

6 0

2 years ago

A piece having a length of 4.0 cm was cut from a much longer, uniform rod. The piece has a volume of 3.0 cm3 and a mass of 24 g.

Gre4nikov [31]

Answer: 96

Explanation:

6 0

2 years ago

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

2 years ago

How often must you check the temperature of food that is being held with temperature control?.

loris [4]

You need to check the temperature of food being stored in a temperature-controlled environment every four hours. The process of changing a space's temperature is called temperature control.

Cooking food alone may not be enough to avoid food poisoning, though, if the bacteria in food are allowed to grow to large numbers. When the temperature is between 5°C and 63°C, bacteria can grow. The risk zone is the range between 5°C and 63°C.

Temperature control is a process where the passage of heat energy into or out of a space or substance is adjusted to achieve the desired temperature. This process involves measuring or otherwise detecting changes in the temperature of the space (and all of the objects contained therein) or of the substance.

Learn more about temperature here

brainly.com/question/11244611

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9 months ago