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3 years ago

I'm new to brainy, but here is my question.

Physics

2 answers:

skad [1K]3 years ago

8 0

680 is the right answer

Fittoniya [83]3 years ago

7 0

Answer: 680 J

Explanation:

Andre's mechanical energy is given by:

$E=K+U$

where

K is the kinetic energy

U is the gravitational potential energy

The kinetic energy in this case is zero, because Andre is sitting, so its speed is zero: v=0, and the kinetic energy is

$K=\frac{1}{2}mv^2=\frac{1}{2}(77.1 kg)(0)^2=0$

The gravitational potential energy is instead

$U=mgh=(77.1 kg)(9.8 m/s^2)(0.90 m)=680.0 J$

So, the total mechanical energy is

$E=U=680.0 J$

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vagabundo [1.1K]

Answer:

The duration of the impact is 0.005384 seconds

Explanation:

Given

m = 0.43 kg

v = 5.2 m/s

x = 0.014 m

Knowing the formulas

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3 years ago

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A charge of 8.5 × 10–6 C is in an electric field that has a strength of 3.2 × 105 N/C. What is the electric force acting on the

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2.72 N

Explanation:

Step 1:

From the basic formula in electrostatics

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E = Electric field strength

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Step 2:

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4 years ago

After asking a question, a scientist can form a(n) , which is an idea that may be proved or disproved by an experiment.

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Answer:

When scientists have a question, they form a hypothesis, which is an idea that may be proved or disproved by an experiment.

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3 years ago

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4 years ago

g A thin-walled hollow cylinder and a solid cylinder, both have same mass 2.0 kg and radius 20 cm, start rolling down from rest

ArbitrLikvidat [17]

Answer:

a. i. 3.43 m/s ii. 2.8 m/s

b. The thin-walled cylinder

Explanation:

a. Find translational speed of each cylinder upon reaching the bottom

The potential energy change of each mass = total kinetic energy gain = translational kinetic energy + rotational kinetic energy

So, mgh = 1/2mv² + 1/2Iω² where m = mass of object = 2.0 kg, g =acceleration due to gravity = 9.8 m/s², h = height of incline = 1.2 m, v = translational velocity of object, I = moment of inertia of object and ω = angular speed = v/r where r = radius of object.

i. translational speed of thin-walled cylinder upon reaching the bottom

So, For the thin-walled cylinder, I = mr², we find its translational velocity, v

So, mgh = 1/2mv² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²)(v/r)²

mgh = 1/2mv² + 1/2mv²

mgh = mv²

v² = gh

v = √gh

v = √(9.8 m/s² × 1.2 m)

v = √(11.76 m²/s²)

v = 3.43 m/s

ii. translational speed of solid cylinder upon reaching the bottom

So, For the solid cylinder, I = mr²/2, we find its translational velocity, v'

So, mgh = 1/2mv'² + 1/2Iω²

mgh = 1/2mv² + 1/2(mr²/2)(v'/r)²

mgh = 1/2mv'² + mv'²

mgh = 3mv'²/2

v'² = 2gh/3

v' = √(2gh/3)

v' = √(2 × 9.8 m/s² × 1.2 m/3)

v' = √(23.52 m²/s²/3)

v' = √(7.84 m²/s²)

v' = 2.8 m/s

b. Determine which cylinder has the greatest translational speed upon reaching the bottom.

Since v = 3.43 m/s > v'= 2.8 m/s,

the thin-walled cylinder has the greatest translational speed upon reaching the bottom.

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3 years ago