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3 years ago

7

Evan drew a diagram to illustrate radiation. What do the arrows represent? gases mixing with particles liquid moving through a v

acuum moving particles electromagnetic waves

2 answers:

sattari [20]3 years ago

7 0

its D. electromagnetic waves.

julsineya [31]3 years ago

4 0

I am pretty sure the answer is d

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What would a force diagram for something WHILE it is being thrown DOWNWARDS look like? <br><br> Ty

Dovator [93]

Answer:

it look the same just to tell you

5 0

3 years ago

An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.

nignag [31]

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

$\epsilon = -L\frac{dI}{dt}$

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

$\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V$

Therefore, the self-induced emf in this inductor is 4.68 mV.

I hope it helps you!

6 0

3 years ago

Pepsi [2]

Answer:

Explanation:

b) Gravity reduces the initial upward velocity to zero in a time of

t = v/g = 40/10 = 4 s

a) h = v₀t + ½gt² = 40(4) + ½(-10)4² = 80 m

or

v² = u² + 2as

h = (0² - 40²) / 2(-10) = 80 m

8 0

2 years ago

What type of data allows geologists to know what rock layers lie underneath Minnesota?

mrs_skeptik [129]

The answer is well log data, it is a detailed log of information taken from a borehole which geologist used to study geological formations of the earth's layer taken from samples returned from the borehole which was dugged.

5 0

3 years ago

A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac

Tpy6a [65]

Answer:

The force on one side of the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

$y=\dfrac{\Delta y}{\sin60}$

$y=\dfrac{2\Delta y}{\sqrt{3}}$

The area of strip is

$dA=\dfrac{9\times2\Delta y}{\sqrt{3}}$

We need to calculate the force on one side of the plate

Using formula of pressure

$P=\dfrac{dF}{dA}$

$dF=P\times dA$

On integrating

$\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}$

$F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}$

$F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})$

$F=3093529.3\ N$

Hence, The force on one side of the plate is 3093529.3 N.

7 0

3 years ago