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Katena32 [7]
3 years ago
11

Some curious students hold a rolling race by rolling four items down a steep hill. The four items are a solid homogeneous sphere

, a thin spherical shell, a solid homogeneous cylinder and a hoop with all its mass concentrated on the hoop's perimeter. All of the objects have the same mass and start from rest. Assume that the objects roll without slipping and that air resistance and rolling resistance are negligible. For each statement below, select True or False.
a) Upon reaching the bottom of the hill, the hoop will have a larger rotational kinetic energy than any of the other objects will when they reach the bottom of the hill.
A: True B: False

b) The hoop reaches the bottom of the hill before the homogeneous cylinder.
A: True B: False

Physics
1 answer:
Papessa [141]3 years ago
8 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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What is gravitational potential energy?
AnnyKZ [126]
I don’t know sorry ok
4 0
3 years ago
If the rotation of a planet of radius 5.32 × 106 m and free-fall acceleration 7.45 m/s 2 increased to the point that the centrip
leonid [27]

Answer:

v = 6295.55 m/s

Explanation:

Given that,

The radius of a planet, r=5.32\times 10^6\ m

The free fall acceleration of the planet, a = 7.45 m/s²

We need to find the tangential speed of a person standing at the equator.

Also, the centripetal acceleration was equal to the gravitational acceleration at the equator.

We know that,

Centri[etal acceleration,

a=\dfrac{v^2}{r}\\\\v=\sqrt{ar}\\\\v=\sqrt{5.32\times10^6\times 7.45}\\\\v=6295.55\ m/s

So, the tangential speed of the person is equal to 6295.55 m/s.

3 0
3 years ago
What is the potential difference per unit length between two infinitely long concentric cylindrical shells with inner radius 1.5
Vinil7 [7]

Answer:

165.8 V/m

Explanation:

The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)

Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)

V = Q/C

V = Q ÷ 2πε₀L/㏑(b/a)

V = Q㏑(b/a)/2πε₀L

So, the potential difference per unit length V' is

V' = V/L = Q㏑(b/a)/2πε₀

Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have

V' = Q㏑(b/a)/2πε₀

V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)

V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)

V' = 0.16575 × 10³ V/m

V' = 165.75 V/m

V' ≅ 165.8 V/m

6 0
3 years ago
Which following options are examples of matter? A) light B) Molecules C) Feelings D) people
yaroslaw [1]
All except C, feelings are not physical.
6 0
4 years ago
Read 2 more answers
Anita holds her physics textbook and complains that it is too heavy. Andrew says that her hand should exert no force on the book
Nady [450]

Answer:

The force exerted is 6189.4 N

Solution:

As per the question:

Area of the book, A = 0.22\times 0.28 = 0.0616\ m^{2}

Mass of the book, m = 3 kg

Pressure of the book, P = 1.0\times10^{5}\ N/m^{2}

Now,

Pressure, P = \frac{F}{A}

Force, F = PA = 1.0\times10^{5}\times 0.0616 = 6160\ N

Force on mass of 3 kg, F' = mg = 3\times 9.8 = 29.4\ N

Now, the force at the bottom is given by:

F'' = F + F' = 6160 + 29.4 = 6189.4 N

5 0
4 years ago
Read 2 more answers
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