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aleksley [76]
3 years ago
14

Which types of electromagnetic waves have frequencies that are higher than those of ultraviolet light but lower than those of ga

mma rays?
microwaves

X-rays

infrared light

radio waves
Physics
2 answers:
yKpoI14uk [10]3 years ago
7 0

The correct answer to the question is X- rays i.e the electromagnetic wave which has frequency that is higher than ultra violet wave and smaller than gamma ray is X-ray.

EXPLANATION:

As per the electromagnetic spectrum of waves, gamma ray has the least wavelength i.e it has the maximum frequency. The minimum frequency that is possessed by electromagnetic wave is radio wave.

If we arrange all the electromagnetic radiations in ascending order of frequency, the perfect order will be -

Radio wave < Micro wave < Infrared wave < Visible light < U.V ray < X - ray < Gamma ray.

From the electromagnetic spectrum, it is obvious that X -rays lie between gamma ray and U.V ray . Its frequency is smaller than gamma ray but higher  than ultra violet wave.

Luba_88 [7]3 years ago
3 0
The answer would be X-Rays.
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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block
borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

T = 2\pi \sqrt{\frac{m}{k} }

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

Now, determine the amplitude of oscillation, A;

E = \frac{1}{2} kA^2

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

Therefore, the amplitude of the oscillation is 2.82 cm

8 0
3 years ago
Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti
Bas_tet [7]

Answer:

v = 0.059 m/s

Explanation:

To find the final speed of Olaf and the ball you use the conservation momentum law. The momentum of Olaf and the ball before catches the ball is the same of the momentum of Olaf and the ball after. Then, you have:

mv_{1i}+Mv_{2i}=(m+M)v  (1)

m: mass of the ball = 0.400kg

M: mass of Olaf = 75.0 kg

v1i: initial velocity of the ball = 11.3m/s

v2i: initial velocity of Olaf = 0m/s

v: final velocity of Olaf and the ball

You solve the equation (1) for v and replace the values of all variables:

v=\frac{mv_{1i}}{m+M}=\frac{(0.400kg)(11.3m/s)}{0.400kg+75.0kg}=0.059\frac{m}{s}

Hence, after Olaf catches the ball, the velocity of Olaf and the ball is 0.059m/s

3 0
3 years ago
Q 28.7: A strip 1.2 mm wide is moving at a speed of 25 cm/s through a uniform magnetic field of 5.6 T. What is the maximum Hall
Zinaida [17]

Answer:the  maximum Hall voltage across the strip= 0.00168 V.

Explanation:

The Hall Voltage is calculated using

Vh= B x v x  w

Where

B is the magnitude of the magnetic field, 5.6 T

v is the speed/  velocity of the strip, = 25 cm/s  to m/s becomes 25/100=0.25m/s

and w is the width of the strip=  1.2 mm to meters becomes 1.2 mm /1000= 0.0012m

 Solving

Vh= 5.6T x 0.25m/s x 0.0012m

=0.00168T.m²/s

=0.00168Wb/s

=0.00168V

Therefore, the  maximum Hall voltage across the strip=0.00168V

3 0
3 years ago
The three types of seismic waves produced by an earthquake are primary, secondary, and
Liono4ka [1.6K]
<span>The three types of seismic waves produced by an earthquake are primary, secondary, and (D.) surface.</span>
7 0
4 years ago
Read 2 more answers
A merry-go-round is shaped like a uniform disk and has moment of inertia of 50,000 kg m 2 . It is rotating so that it has an ang
lapo4ka [179]

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

<u>r = 20 m</u>

5 0
3 years ago
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