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3 years ago

Why is the ocean warmer than the air at night

1 answer:

4 0

You see, during the day the ocean collects heat from the sun. So the air above the ocean get warm at night, but the rest of the air on the land gets cooler because water has the ability to collect energy from the Sun.

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Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during

suter [353]

Answer: Option a.

Explanation:

Kepler's 2nd law of planetary motion states:

A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

3 0

3 years ago

What is a concave lens?

ozzi

A concave lens is a lens that has at least one of its surfaces or both surfaces curved inwards. Due to this reason, this lens diverges the light that falls on it and hence is also called a diverging lens. The concave lens is thinner in the middle compared to its edges. These are used in flashlights, binoculars, telescopes, etc.
Please see attached image for reference.

5 0

3 years ago

Which particle is commonly used to initiate a fission chain reaction

Snezhnost [94]

Neutron is commonly used to initiate a fission chain reaction.

5 0

3 years ago

A. According to theory, the period T of a simple pendulum is T = 2????√ ???? ???? a. If ???? is measured as ???? = 1.40 ± 0.01 m

salantis [7]

Answer:

a) T = (2,375 ± 0.008) s , b) When comparing this interval with the experimental value we see that it is within the possible theoretical values.

Explanation:

a) The period of a simple pendulum is

T = 2π √ L / g

Let's calculate

T = 2π √1.40 / 9.8

T = 2.3748 s

The uncertainty of the period is

ΔT = dT / dL ΔL

ΔT = 2π ½ √g/L 1/g ΔL

ΔT = π/g √g/L ΔL

ΔT = π/9.8 √9.8/1.4 0.01

ΔT = 0.008 s

The result for the period is

T = (2,375 ± 0.008) s

b) the experimental measure was T = 2.39 s ± 0.01 s

The theoretical value is comprised in a range of [2,367, 2,387] when we approximate this measure according to the significant figures the interval remains [2,37, 2,39].

When comparing this interval with the experimental value we see that it is within the possible theoretical values.

6 0

4 years ago

Rank the angular speed of the following objects, from highest to lowest:- A bowling ball of radius 12.3 cm rotating at 8.21 radi

dlinn [17]

Answer:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

Explanation:

To rank the angular speed (ω) of the objects, we need first calculate its value for every object:

A bowling ball of radius 12.3cm rotating at 8.21 radians per second:

ω = 8.21 rad/s

A tire of radius 0.321m rotating at 75.8 rpm:

$\omega = 75.8 \frac{rev}{min}\cdot \frac{2\pi rad}{1rev}\cdot \frac{1min}{60s} = 7.94rad/s$

A 6.84cm diameter top spinning at 375 degrees per second:

$\omega = 375 \frac{^\circ}{s}\cdot \frac{2\pi rad}{360^ \circ} = 6.54rad/s$

A square with sides (b) 0.458m long, whose corners are moving with tangential speed (v) 2.51 m/s as it rotates about its center:

$\omega = \frac{v}{r} = \frac{v}{\frac{b}{2}\sqrt 2} = \frac{2.51 m/s}{\frac{0.458 m}{2} \sqrt 2} = 7.75rad/s$

A rock on a string, being swung in a circle of radius 0.521 m with a centripetal acceleration (a) of 25.4 m/s²:

$\omega = \sqrt \frac{a}{r} = \sqrt \frac{25.4 m/s^{2}}{0.521m} = 6.98rad/s$

Now, the rank of the angular speed of the objects, from highest to lowest is:

A bowling ball (8.21rad/s) > A tire (7.94rad/s) > A square (7.75rad/s) > A rock (6.98rad/s) > A top spinning (6.54rad/s)

I hope it helps you!

4 0

3 years ago