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loris [4]
3 years ago
10

6. How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?

Physics
1 answer:
Zina [86]3 years ago
8 0

Answer:

60 W

Explanation:

The work done in lifting the object is given by:

W=Fd

where

F = 30.0 N is the weight of the object

d = 10.0 m is the displacement covered by the object

Using the formula,

W=(30.0 N)(10.0 m)=300 J

Now we can calculate the power, which is given by

P=\frac{W}{t}

where

W = 300 J is the work done

t = 5.00 s is the time interval

Substituting,

P=\frac{300 J}{5 s}=60 W

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How many protons are in this atom if it has a balanced charge? 0 2 4 6
pickupchik [31]

Answer:

For an atom to have a balanced charged, the number of protons shall be equal to the number of electrons of an atom. proton is the positive part of an atom whereas electrons are the negative part of an atom. Only if the number of protons will be equal to the number of electron, the atom will be able to be neutral.

If the number of electron will be more, then the atom will be negative. If the number of electron will be less, then the atom will be positive.

7 0
3 years ago
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A 4.0-kg object is supported by an aluminum wire of length 2.0 m and diameter 2.0 mm. How much will the wire stretch?
forsale [732]

Answer:

The extension of the wire is 0.362 mm.

Explanation:

Given;

mass of the object, m = 4.0 kg

length of the aluminum wire, L = 2.0 m

diameter of the wire, d = 2.0 mm

radius of the wire, r = d/2 = 1.0 mm = 0.001 m

The area of the wire is given by;

A = πr²

A = π(0.001)² = 3.142 x 10⁻⁶ m²

The downward force of the object on the wire is given by;

F = mg

F = 4 x 9.8 = 39.2 N

The Young's modulus of aluminum is given by;

Y = \frac{stress}{strain}\\\\Y = \frac{F/A}{e/L}\\\\Y = \frac{FL}{Ae} \\\\e = \frac{FL}{AY}

Where;

Young's modulus of elasticity of aluminum = 69 x 10⁹ N/m²

e = \frac{FL}{AY} \\\\e = \frac{(39.2)(2)}{(3.142*10^{-6})(69*10^9)} \\\\e = 0.000362 \ m\\\\e = 0.362 \ mm

Therefore, the extension of the wire is 0.362 mm.

8 0
3 years ago
A rotating object starts from rest at t = 0 s and has a constant angular acceleration. At a time of t = 2.5 s the object has an
Evgesh-ka [11]

Answer:

52 rad

Explanation:

Using

Ф = ω't +1/2αt²................... Equation 1

Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.

Since the object states from rest, ω' = 0 rad/s.

Therefore,

Ф = 1/2αt²................ Equation 2

make α the subject of the equation

α = 2Ф/t².................. Equation 3

Given: Ф = 13 rad, t = 2.5 s

Substitute into equation 3

α = 2(13)/2.5²

α = 26/2.5

α = 4.16 rad/s².

using equation 2,

Ф = 1/2αt²

Given: t = 5 s, α = 4.16 rad/s²

Substitute into equation 2

Ф = 1/2(4.16)(5²)

Ф = 52 rad.

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A ball is thrown at an angle of 38° to the horizontal. What happens to the
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Answer:

Vy = V0 sin 38       where Vy is the initial vertical velocity

The ball will accelerate downwards (until it lands)

Note the signs involved   if Vy is positive then g must be negative

The acceleration is constant until the ball lands

t (upwards) = (0 - Vy) / -g      = Vy / g      final velocity = 0

t(downwards = (-Vy - 0) / -g = Vy / g      final velocity = -Vy

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8 0
3 years ago
Which example show harassment?
koban [17]

C. making fun of a peer because she is Asian

hope this helps

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3 years ago
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