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3 years ago

8

Why do sea breezes occur during the day?

2 answers:

Sergio039 [100]3 years ago

7 0

Sea breezes occur during the day because solar radiation warms the land more than the water. This causes the warmer air over the land to rise. The resulting convection current causes wind to blow in from the sea.

Hatshy [7]3 years ago

6 0

The reason why is because the solar radiation warms land more than wind and because the unequal heating rates of land and water. Hope this helps :)

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Please answer these two questions ASAP!! Will mark brainliest and give points!

maw [93]

Counter clockwise torque is 360Nm.
Clockwise torque is 240Nm.

40 * 9 = 360
80 * 3 = 240

6 0

3 years ago

Read 2 more answers

A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85

artcher [175]

Answer:

$v = 10.75\,\frac{m}{s}$

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

$(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )$

The velocity of the bowling ball after the collision is:

$v = 10.75\,\frac{m}{s}$

8 0

3 years ago

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Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

<u>So, the heat rate:</u>

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

<u>From the Fourier's law of conduction we have:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

How close would you have to bring 1 C of positive chargeand 1 C of negative charge for them to exert forces of 1 N onone another

Harman [31]

Answer:

94,800 m

Explanation:

F = kq1 q2/r^2

1 = 9 x 10^9 x 1 / r^2

4 0

2 years ago

At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?

Ipatiy [6.2K]

The resistance of the sample is $682.5\Omega$

Explanation:

The relationship between resistance of a material and temperature is given by the equation

$R(T)=R_0(1+\alpha (T-T_0))$

where

$R_0$ is the resistance at the temperature $T_0$

$\alpha$ is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

$R_0 = 525 \Omega$ when the temperature is $T_0 = 20^{\circ}C$

While the temperature coefficient of resistance of nickel is

$\alpha = 0.006/^{\circ}C$

Therefore, the resistance of the sample when its temperature is

$T=70^{\circ}C$

is

$R=(525)(1+0.006(70-20))=682.5 \Omega$

Learn more about resistance:

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brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

3 0

3 years ago