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stealth61 [152]
3 years ago
11

At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people o

n the equator were just 'thrown off'? Why is the expression 'thrown off' a bad one ?
Physics
2 answers:
Elina [12.6K]3 years ago
6 0
Hello there.
<span>
At ground level g is 9.8m/s^2. Suppose the earth started to increase its angular velocity. How long would a day be when people on the equator were just 'thrown off'? Why is the expression 'thrown off' a bad one ?
</span>
People would not be thrown off.

velikii [3]3 years ago
5 0
Let s = rate of rotation 
<span>Let r = radius of earth = 6,400km </span>
<span>Then solving (s^2) r = g will give the desired rate, from which length of day is inferred. </span>
<span>People would not be thrown off. They would simply move eastward in a straight line while the curved surface of earth fell away from beneath them.</span>
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The figure shows a 100-kg block being released from rest from a height of 1.0 m. It then takes it 0.90 s to reach the floor. Wha
Anna007 [38]

Answer:

The mass of the another block is 60 kg.

Explanation:

Given that,

Mass of block M= 100 kg

Height = 1.0 m

Time = 0.90 s

Let the mass of the other block is m.

We need to calculate the acceleration of each block

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

1.0=0+\dfrac{1}{2}\times a\times(0.90)^2

a=\dfrac{2\times1.0}{(0.90)^2}

a=2.46\ m/s^2

We need to calculate the mass of the other block

Using newton's second law

The net force of the block M

Ma=Mg-T

T=Mg-Ma....(I)

The net force of the block m

ma=T-mg

Put the value of T from equation (I)

ma=Mg-Ma-mg

m(a+g)=M(g-a)

m=\dfrac{M(g-a)}{(a+g)}

Put the value into the formula

m=\dfrac{100(9.8-2.46)}{2.46+9.8}

m=59.8\ \approx60\ kg

Hence, The mass of the another block is 60 kg.

8 0
3 years ago
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un
Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = Iw = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

(I_{1} +I_{2} )w

where the subscripts 1 and 2 indicates the values first and second  flywheels

(I_{1} +I_{2} )w = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = \frac{1}{2}mv_{car} ^{2}

where m is the mass of the car

v_{car} is the velocity of the car

Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

3 0
3 years ago
If the voltage drop across the first resistor is 13.00V, then how many volts remain for the 2nd resistor?
Scorpion4ik [409]

Answer:

2+1

Explanation:

2+1

8 0
2 years ago
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When you go camping, you burn wood. Are you contributing to air pollution?
Jet001 [13]
Yes because of the smoke you are creating in the air
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3 years ago
Question 2 of 34
tatiyna

You can find the magnitude of the resultant vector : (B). By adding the magnitudes of the two vectors

<h3>Meaning of Vectors</h3>

A vector can be defined as any quantity which possesses magnitude and also has direction.

A Vector quantity is very useful because This type of quantity gives more details to the student or teacher analyzing it.

Lear more about Vectors: brainly.com/question/25705666

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2 years ago
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