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4 years ago

Which types of electromagnetic waves have higher frequencies then the waves that make up ultraviolet light?

2 answers:

5 0

Answer: Visible light makes up just a small part of the full electromagnetic spectrum. Electromagnetic waves with shorter wavelengths and higher frequencies include ultraviolet light, X-rays, and gamma rays.

vredina [299]4 years ago

5 0

Answer:

The answer is D

Explanation:

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In flight, a rocket is subjected to four forces; weight, thrust, lift, and drag. Forces are vector quantities that have both a m

Flura [38]

Answer:

thrust

Explanation:

4 0

3 years ago

Read 2 more answers

Identification of elements compounds and mixtures from a given table

Firdavs [7]

Answer:

the elements are made up of only one kind of atom, the compounds consist of two or more elements that are chemically combined, and the mixtures are physical combination of two or more substances

3 0

3 years ago

Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that

kaheart [24]

Answer:

Young's modulus for the rope material is 20.8 MPa.

Explanation:

The Young's modulus is given by:

$E = \frac{FL_{0}}{A\Delta L}$

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

$A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}$

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

$F = T_{w} = W_{o}$

Where:

$T_{w}$ : is the tension of the wire

$W_{o}$ : is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

$F = mg = 1700 kg*9.81 m/s^{2} = 16677 N$

Hence, the Young's modulus is:

$E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa$

Therefore, Young's modulus for the rope material is 20.8 MPa.

I hope it helps you!

7 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s

Let $S_{0}$ be the speed of the river's current given as 1.00 m/s.

Note that this speed is the magnitude of the velocity which is a vector quantity.

The direction of the swimmer is upstream.

Hence the resultant velocity is given as, $S_{R}$ = S — S 0 $S_{0}$

$S_{R}$ = 1.25 — 1

$S_{R}$ = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0

4 years ago