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geniusboy [140]

3 years ago

8

A woman on a 10-speed bicycle travels at 9m/s relative to the ground as she passes a little boy on a tricycle going in the oppos

ite direction. If the boy is travelling at 1m/s relative to the ground , how fast does the boy appear to be moving relative to the woman

1 answer:

Archy [21]3 years ago

8 0

Answer:

23

Explanation:

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The process of hotter air rising and cooler air falling is called convection. What role does convection play in the atmosphere?

erastovalidia [21]

Option B will be the right one !

7 0

4 years ago

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Two coils that are separated by a distance equal to their radius and that carry equal currents such that their axial fields add

jok3333 [9.3K]

Answer:

When x = 2.8 cm, $B_{x1} = 0.0265 T$

When x = 5.5 cm, $B_{x2} = 0.0209 T$

when x = 7.3 cm, $B_{x3} = 0.0169 T$

When x = 11.0 cm, $B_{x4} = 0.0103 T$

Explanation:

According to Biot-Savart law,

$B_{x} = \frac{N \mu_{o}IR^{2} }{2(x^{2} +R^{2} )^{3/2} }\\$ .......................(1)

R = 11.0 cm = 0.11 m

I = 17.0 A

N = 300 turns

$\mu_{o} = 4\pi * 10^{-7} N/A^{2}$

When x₁ = 2.8 cm = 0.028 m

$B_{x1} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.028^{2} +0.11^{2} )^{3/2} }\\B_{x1} = 0.0265 T$

When x₂ = 5.5cm = 0.055 m

$B_{x2} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.055^{2} +0.11^{2} )^{3/2} }\\B_{x2} = 0.0209 T$

When x₃ = 7.3 cm = 0.073 m

$B_{x3} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.073^{2} +0.11^{2} )^{3/2} }\\B_{x3} = 0.0169 T$

When X₄ = 11.0 cm = 0.11 m

$B_{x4} = \frac{300 *(4\pi * 10^{-7} ) * 17 *0.11^{2} }{2(0.11^{2} +0.11^{2} )^{3/2} }\\B_{x4} = 0.0103 T$

4 0

4 years ago

Can someone plz help me no links plz

Orlov [11]

Answer:

Im doing edge too rn-

Explanation:

Can you tell me the drop down options

7 0

3 years ago

Suppose a spacecraft orbits the moon in a very low, circular orbit, just a few hundred meters above the lunar surface. The moon

Finger [1]

Spacecraft will be moving in 1700 m/s.

Option C

<h3><u>Explanation: </u></h3>

The diameter of the moon is 3500 km and the free fall acceleration at the surface is given as $1.6\ \mathrm{m} / \mathrm{s}^2$

The radius will be half of the diameter of the moon that can be written as:

$r_{\text {moon }}=1.75 \times 10^{6}$

By the application of the equation for orbit speed, we get

$\begin{aligned}&v_{\text {orbit}}=\sqrt{r \times q}\\&v_{\text {orbit}}=\sqrt{\left(1.75 \times 10^{6}\right) \times\left(1.6\ \mathrm{m} / \mathrm{s}^{2}\right)}\\&v_{\text {orbit}}=1700\ \mathrm{m} / \mathrm{s}\end{aligned}$

5 0

3 years ago

The momentum of an object depends on what two forces

mr_godi [17]

The Linear momentum depends on

1) Mass

2) Linear Velocity

The angular momentum depends on

1) Inertia

2) Rotational velocity

7 0

3 years ago