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2 years ago

6

A 2 kg apple is at rest on the floor at the bottom of a staircase. How much work would be necessary to bring that apple up to th

e top of the staircase which is 8 meters high?
Answers:
16.0 J
160 J
64.0 J
640

2 answers:

vodka [1.7K]2 years ago

7 0

700.5 J I think this is the answer

maw [93]2 years ago

3 0

Answer:

a

Explanation:

16.0 J literally just doubled the number

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A car traveling on a straight road at 10 meters per second accelerates uniformly to a speed of 21 meters per second in 12 second

labwork [276]

21+10=31 because you can see that 21 and 10 are in metres while 12 is in seconds so 21+10=31 is the answer.

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2 years ago

If the car is traveling at a velocity of 15 m/s, what is the approximate centripetal acceleration of the car?.

BaLLatris [955]

The vehicle's centripetal acceleration is equal to 22.5m/s²

Radius, r = 10 meter

Speed, V = 15 m/s

To ascertain the car's centripetal acceleration

A(c) = V²/R

We obtain the following when we enter the formula's parameters:

A(c) = 152/10

A(c) = 225/10

A(c) = 22.5m/s²

<h3>What is Centripetal acceleration ?</h3>

When an item moves in a circular route, one of its motion characteristics is centripetal acceleration. Any motion in a circle with an acceleration vector pointing in the direction of the circle's centre is referred to as centripetal acceleration.

Centripetal forces cause accelerations at the centripetal axis. With the exception of the Earth's rotation around the Sun, any satellite's circular motion around a celestial body is brought on by the centripetal force produced by their mutual gravitational pull.

Hence, Centripetal acceleration is

22.5 m/s²

Learn more about Centripetal acceleration here:

brainly.com/question/79801

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7 0

11 months ago

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

xxMikexx [17]

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$ . Hence, this is the required solution.

3 0

2 years ago

Read 2 more answers

Which statements apply to transverse waves? Check all that apply.

Levart [38]

Correct Answers:

- The waves have a trough

- the waves have a crest

- the energy that is transferred move in a perpendicular direction

- particles move in an up and down motion

8 0

2 years ago

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Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par

Murrr4er [49]

Answer:

X₃₁ = 0.58 m and X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

F₁₃ - F₂₃ = 0

F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

F₁₃ = k q₁ q₃ / r₁₃²

F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

r₁₃ = x₃- 0

r₂₃ = 2 –x₃

We substitute

k q q / x₃² = k 4q q / (2-x₃)²

q² (2 - x₃)² = 4 q² x₃²

4- 4x₃ + x₃² = 4 x₃²

5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5

x₃ = [-4 ± 9.80] 10

X₃₁ = 0.58 m

X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

7 0

2 years ago