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2 years ago

8

The radius of the earth is 6.37 x 10ºm How fast in meters per second is a tree on the equator moving because of the earth's rota

tion? How fast is a polar bear at the north
pole?

1 answer:

REY [17]2 years ago

3 0

Answer:

yea what the other guy said

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A person who studies physics is known as

Leto [7]

Physicists. Welcome :)

4 0

2 years ago

Read 2 more answers

A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of

adelina 88 [10]

Answer:

x = 11.23 m

Explanation:

For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.

Let's reduce to SI system units

θ = 155 rev (2pi rad / rev) = 310π rad

α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²

Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)

w² = w₀² + 2 α θ

w =√ 2 α θ

w = √(2 4pi 310pi)

w = 156.45 rad / s

The relationship between angular and linear velocity

v = w r

v = 156.45 0.175

v = 27.38 m / s

In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive

y = $v_{oy}$ t - ½ g t²

As it leaves the highest point its speed is horizontal

y = 0 - ½ g t²

t = √ (-2y / g)

t = √ (-2 (-0.820) /9.8)

t = 0.41 s

With this time we calculate the horizontal distance, because the constant horizontal speed

x = vox t

x = 27.38 0.41

x = 11.23 m

5 0

3 years ago

I drove from Oklahoma to California (1452.9 miles) in 23 hours? What was my average speed?

storchak [24]

Answer:

63 miles per hour

Explanation:

you just divide miles and hours to get your average speed

5 0

2 years ago

A 1502.7 kg car is traveling at 33.1 m/s when

Bond [772]

By definition of average acceleration,

<em>a</em> = (20 m/s - 33.1 m/s) / (4.7 s) ≈ -2.78 m/s²

Vertically, the car is in equilibrium, so the net force is equal to the friction force in the direction opposite the car's motion:

∑ <em>F</em> = (1502.7 kg) (-2.78 m/s²) ≈ -4188.38 N ≈ -4200 N

If you just want the magnitude, drop the negative sign.

5 0

2 years ago

Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le

Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = $\frac{d}{2}$

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

$L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s$

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

$K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J$

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

$L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s$

Energy

$E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J$

The new energy will be 1000 J

Work done will be the change in the kinetic energy

$W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J$

The work done is 600 J

5 0

2 years ago