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2 years ago

5

Tropical winds that blow east to west are?

1 answer:

suter [353]2 years ago

6 0

Its called the "westerly"

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When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If

Alex787 [66]

Answer:

a) y = 0.0075 m

b) W = 1.569 J

Explanation:

See attachment for the solution

6 0

2 years ago

Question 13 of 20

Dominik [7]

Answer:

B. It has a central nucleus composed of 29 protons and 35 neutrons,

surrounded by an electron cloud containing 29 electrons.

Explanation:

Protons and neutrons are the only subatomic particles with mass, and they are located in the nucleus. If this atom has an atomic number of 29 and is copper, it must have 29 protons (protons define which element is being observed). This means all remaining mass is from neutrons. 64-29 = 35.

Electrons have no mass and orbit the nucleus in the electron cloud. Since this copper atom is neutral (we are not told it has a charge), there must be an equal number of protons and electrons.

6 0

2 years ago

Read 2 more answers

An insulating sphere is 8.00 cm in diameter and carries a 6.50 ÂµC charge uniformly distributed throughout its interior volume.

Kobotan [32]

Explanation:

(a) Formula to calculate the density is as follows.

$\rho = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}$

= $2.42 \times 10^{-2} C/m^{3}$

Now, calculate the charge as follows.

$q_{in} = \rho(\frac{4}{3} \pi r^{3})$

= $2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}$

= $10.106 \times 10^{-8}$ C

or, = 101.06 nC

(b) For r = 6.50 cm, the value of charge will be calculated as follows.

$q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}$

= $\frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}$

= 7.454 $\mu C$

7 0

2 years ago

describe the time of day that an early explorer might have planned to enter a harbor and when he might have planned to leave

inessss [21]

He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.

4 0

2 years ago

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0

2 years ago