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3 years ago

14

A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from

its starting point?

1 answer:

Alla [95]3 years ago

6 0

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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A light bulb radiates 110 nW of single-frequency sinusoidal electromagnetic waves uniformly in all directions. Calculate the ave

krek1111 [17]

Answer:

The intensity of the light from the bulb would be

3.501 x $10^{-6}$ W/ $m^{2}$

Explanation:

Given

The Power = 110 n W = 110 x $10^{-9}$ W

the distance r = 50 mm = 50 /1000 = 0.05 m

The intensity can be obtained with the relationship below;

I = Power/area ......1

The area of the sphere would be used in this case since the bulb is spherical; A=4π $r^{2}$

Putting it into equation 1, we have;

I = P/ 4π $r^{2}$

I = 110 x $10^{-9}$ / 4 x π x $0.05^{2}$

I = 3.501 x $10^{-6}$ W/ $m^{2}$

Therefore the intensity of the light from the bulb would be

3.501 x $10^{-6}$ W/ $m^{2}$

6 0

3 years ago

The stopping distance d of a car after the brakes are applied varies directly as the square of the speed r. If a car travelling

Crank

270/70^2 = x/80^2

Cross multiply

270 (6400) = 4900x

x = 270(6400)/4900

352 and 32/49 feet

Hope this helps

4 0

3 years ago

If a car has a mass of 200 kilograms and produces a force of 500 N, how fast will the car accelerate?

san4es73 [151]

Answer:

B

Explanation:

F = ma

500N = 200 × a

200a. = 500N

a. =. 500N/200

a. =. 2.5m/s²

PLEASE GIVE BRAINLIEST.

3 0

2 years ago

_____ friction is the force that sliding objects experience

Gnesinka [82]

Answer:

Sliding friction is the force that sliding objects experience

Explanation:

7 0

3 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago