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3 years ago

If you move a force of 30.0 N a distance of 2.00 m, what is the amount of work done?

Physics

1 answer:

vladimir1956 [14]3 years ago

6 0

Answer:

A. 60 J

Explanation:

Use the equation $W=Fd$ | F= force, D= distance.

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Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the

wariber [46]

First, find the amount of time for the dart to hit the board using this equation: t = d/v

t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>

3 0

3 years ago

The distance between two stations is 1995 Km. How much time will it take to cover the distance at an average speed of 19KM/hour

Flura [38]

105 hours or 4.375 days

5 0

3 years ago

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

Explain a situation in which you can accelerate even though your speed doesn’t change.

Serga [27]

"Acceleration" does NOT mean speeding up. It also doesn't mean
slowing down. Acceleration means ANY change in the speed
OR DIRECTION of motion.

The only kind of motion that's NOT accelerated is motion at a steady
speed AND in a straight line.

Even when your speed is steady, you're accelerating if your direction
is changing.

A few examples:
(no speeds are changing):

-- driving on a curved road, or turning a corner
-- going around a curve on a skateboard, a bike, or a Segway
-- running on a quarter-mile track
-- an Indy car cruising a practice lap around the track
-- water spinning, getting ready to go down the drain
-- any point on the blade of a fan
-- the little ball going around the inside of a Roulette wheel
-- the Moon in its orbit around the Earth
-- the Earth in its orbit around the sun

4 0

3 years ago

Light from a single laser is directed through two slits that are separated by a small distance. On the other side of the slits i

nekit [7.7K]

The answer is destructive interference. You have this for both C and D. I suspect one of C or D is supposed to be constructive interference... But destructive interference is the answer

3 0

3 years ago