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3 years ago

5

Due to the earth's rotation, a person would be traveling faster at the equator than they would at a position halfway from the eq

uator to the North Pole.

2 answers:

Kisachek [45]3 years ago

8 0

Answer:

False

Explanation:

It is actually the exact opposite.

Vinil7 [7]3 years ago

3 0

Answer:

Explanation:

Yes, If tangential velocity is your reference frame

If R is the radius of earth

and ω is the angular velocity = (2π / (24(3600)) radians/s

Tangential velocity at the equator is

v = ωR = (2π / (24(3600))(6 370 000 m) = 463 m/s

Tangential velocity at a point half way to the pole (45° latitude)

v = ωRcosΘ = 463cos45 = 328 m/s

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In which direction does the electric field point at a position directly east of a

svlad2 [7]

Answer:

<h2>D. East </h2><h3>That's my correct answer</h3>

3 0

2 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the

uysha [10]

Answer:

3.14 × 10⁻⁴ m³ /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as

d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.

Speed in the first segment of the pipe is

v1 = 4 m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow rate = Q = A1 v1 = A2 v2 = A3 v3.

Q = A1v1

=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s

3 0

3 years ago

nikitadnepr [17]

The answer is
B:because

8 0

3 years ago

When is a white dwarf formed?

Anna [14]

When a red giant has insufficient mass it pretty much sheds its outer layer to leave the center core know as the white dwarf

6 0

3 years ago

Read 2 more answers