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3 years ago

Please help me! Science is not my best subject

Physics

1 answer:

Paul [167]3 years ago

8 0

1)solid
2)contact
3)conduction
4)Fluid
5)Less
6)More
7)Convection
8)Waves
9)Radiation
10)emit
11)absorb

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Which energy source is formed when organic matter is trapped underground without exposure to air or moisture?

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C. Coal Is The Answer

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The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove

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Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, $d = \frac{m \lambda}{sin(A_{m}) }$ ..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, $\lambda$ = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

$A_{m}$ = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

$sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}$

$sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}$

Putting all appropriate values into equation (1)

$d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m$

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3 years ago

the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass

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That is very true, but what is the question asking you.

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During a 980. miles cruise, a ship travelled 590 miles in 7 hours first. Then the ship changed its speed for the rest of the tri

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A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

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Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
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So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

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