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schepotkina [342]
3 years ago
5

Determine the mechanical energy of this object a 1-kg ball rolls on the ground at m/s

Physics
1 answer:
dedylja [7]3 years ago
8 0
Mechanical energy = potential energy + kinetic energy
The ball is on the ground so it has no potential energy. that's all i know.
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A 125-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a
lbvjy [14]

Answer:

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

4 0
3 years ago
What property do liquids and gases share
vodka [1.7K]

Answer:

A is correct

Explanation:

6 0
3 years ago
Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod
Dima020 [189]

Answer:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>

Explanation:

If the temperature of  the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.

<u>We can prove this by the equation of heat for the two bodies:</u>

<em>According to given condition,</em>

\Delta T_1=\Delta T_2

\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}

<em>when there is no heat loss from the system of two bodies then </em>Q_1=Q_2

\frac{1}{m.c_1} =\frac{1}{m.c_2}

\Rightarrow c_1=c_2

  • Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

  • when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

OR

  • the mass of colder object is half the mass of the hotter object while their specific heat is same.
3 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M_S = 1.99 × 10³⁰ kg

Mass of the neutron star

M_N = 2( M_S )

M_N = 2( 1.99 × 10³⁰ kg )

M_N = ( 3.98 × 10³⁰ kg )

Radius of neutron star R_N = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω_N.

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM_N = / R_N² = mR_Nω_N²

ω_N² = GM_N = / R_N³

ω_N = √(GM_N = / R_N³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω_N = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω_N = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω_N = √ 120831133.3636777

ω_N = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0
3 years ago
What were the two different alleles for height in the pea plants that Mendel studied?
natta225 [31]
You might want to go to this website,       http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P
8 0
3 years ago
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