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DiKsa [7]
3 years ago
14

If an object has a mass of 20 grams and a volume of 40 cm3, what is its density in g/cm3?

Physics
1 answer:
alexgriva [62]3 years ago
8 0
20/40=0.5 g/cm^3 becuase, mass/volume=density.
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Neurons in our bodies carry weak currents that produce detectable magnetic fields. A technique called magnetoencephalography, or
Stolb23 [73]

Answer:

I = (1.80 × 10⁻¹⁰) A

Explanation:

From Biot Savart's law, the magnetic field formula is given as

B = (μ₀I)/(2πr)

B = magnetic field = (1.0 × 10⁻¹⁵) T

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

r = 3.6 cm = 0.036 m

(1.0 × 10⁻¹⁵) = (4π × 10⁻⁷ × I)/(2π × 0.036)

4π × 10⁻⁷ × I = 1.0 × 10⁻¹⁵ × 2π × 0.036

I = (1.80 × 10⁻¹⁰) A

Hope this Helps!!!

3 0
3 years ago
Mirage is due to...
Irina18 [472]
<span>Mirage is due to (A). unequal heating of different parts of the atmosphere,
because that produces unequal density of different parts of the atmosphere,
which bends the light.
</span>
3 0
3 years ago
Describe what the effect of increasing the power of a camera would have on the battery life
patriot [66]

Answer:

. Cut Down on the LCD

The biggest battery drain in a camera is the LCD – both the rear screen and the electronic viewfinder. This is the big reason why DSLRs almost always have longer battery life specifications than mirrorless cameras – the optical viewfinder lets you skip LCDs altogether.

However, if you use your DSLR in live view, you’ll notice that its battery life slides dramatically. Side by side against a mirrorless camera, there’s actually a good chance it will die first. LCDs just take a lot of power to run.

What does this imply? Quite simply, you should always do what you can to cut down on LCD usage when your battery is running low.

For DSLR users, that means switching to the optical viewfinder. For mirrorless photographers, it means turning off the camera frequently, or setting it so the viewfinder only activates when you hold it to your eye.

And regardless of the camera you use, drastically cut down on the amount of time you spend reviewing photos. Chimping has its place, but not while your battery warning is blinking red.Optimize Your Battery Saver Settings

Most cameras have menu options designed to improve battery life and maximize your shooting time. For example, the “metering timeout” setting lets you select how long you want the camera to wait during inactivity before shutting off its metering system.

Beyond that, a number of cameras today have an “Eco mode” that minimizes power consumption from the camera’s LCD. On the Canon EOS R, for example, Eco mode dims and then turns off the LCD when not in use, improving your battery life significantly – from 370 to 540 shots per charge, according to Canon’s official specifications.

It’s also important to note that mirrorless cameras are generally more efficient using the rear LCD than the electronic viewfinder. In terms of the EOS R again, Canon only rates 350 shots using the EVF, with no Eco mode to improve it. On the Sony side of things, the new A7R IV is rated for 530 shots via the viewfinder and 670 via the rear LCD.

If none of that applies to you, one option at your disposal is always to lower the brightness of your rear LCD. It might make photography a bit trickier in bright conditions, but the payoff is getting the shot rather than missing it completely due to a dead battery.

Other camera settings and extras that harm battery life include:

Image stabilization (both in-body and in-lens)

Popup flash

Bluetooth and WiFi

Most external accessories: GPS dongles, lightning triggers, wireless remote releases, shotgun mics, etc.

Sometimes, these capabilities are essential for your photo, so it’s worth the battery life sacrifice. But if you’re down to your last bar, double check to ensure that you’re not using any of the above settings or accessories without good reason.

8 0
3 years ago
A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.
kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0
3 years ago
A distance of 1.0 meter separates the centers of
professor190 [17]
Neither set of choices is correct.

If the distance is tripled, then the forces decrease to

1/9 Fg. and. 1/9 Fe.

Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.
4 0
3 years ago
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