A lightbulb has a power rating of 6.75 W and draws a current of 0.75 A when connected to a battery.

A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional f

drek231 [11]

Answer:

Explanation:

According to energy conservation which states that the workdone is equal to change in the system

Workdone = change in kinetic energy + (frictional force * distance)

Workdone = ΔK + fd

Workdone = kf-Ki + fd

Workdone = = 1/2(m(v-u)^2) + fd

Given

Mass m = 495kg

final velocity v = 105m/s

initial velocity = 0m/s

Force f= 1400N

distance d = 395m

Substitute

Workdone = 1/2(495(105-0)^2) + 1400(395)

Workdone = 2,728,687.5+553000

Workdone = 3,281,687.5 Joules

Time = 8.2secs

Power output = Workdone/Time

Power output = 3,281,687.5/8.2

Power output = 885,766.768

Power output = 8.858 * 10^5 watts

3 0

4 years ago

Plz help i really need this

ddd [48]

Answer:

The reactivity of metals increases as you move left in a period and as you move down in a group, so Marie needs to know the period and group of the element inside each box. Boxes that show locations in groups 1 or 2 or in period 8 contain the most reactive elements.

Explanation:

6 0

3 years ago

Read 2 more answers

Usain Bolt, a Jamaican sprinter, holds the Olympic and world records for the 100-m and 200-m dash, which he

stellarik [79]

Answer: 10.36m/s

How? Just divide 200m by 19.3 and you will get how fast he ran per m/s

6 0

2 years ago

Sherlock Holmes examines a clue by holding his magnifying glass (with

Alborosie

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

$\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm$

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

$M=-\frac{q}{p}=-\frac{-30}{10}=3$

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

8 0

4 years ago

Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

marin [14]

Answer:

Part a)

$Q = 952 cal/day$