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insens350 [35]
3 years ago
12

Which part of the electromagnetic Spectrum is nearest to X-rays?

Physics
1 answer:
den301095 [7]3 years ago
7 0

Answer:gamma rays

Explanation:

Out of all these spectrum listed in The question , gamma rays is closest to x-ray in The electromagnetic spectrum

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When there are two light sources then there will be two shadows. Given below is a case where there are two light sources but one
velikii [3]

neither, a shadow cant be darker unless your closer to the surface your pointing the light at.

3 0
3 years ago
how much time would it take for the sound of a thunder to travel 1,500 meters if sound travelers at a speed of 330 m/sec?
inysia [295]

(1,500 meters) x (1 sec/330 meters) =

(1,500 / 330) (meters-sec/meters) =

4.55 seconds

8 0
3 years ago
A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s
Evgen [1.6K]

Answer:

(a) I (Moment of inertia)=0.0987 kgm^{2}

(b) W(Angular Speed)=2.66 \frac{rad}{s}

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

T=2(3.14)\sqrt{\frac{I}{mgd} }

After Simplifying we get

I=\frac{mgdT^{2} }{4*(3.14)^{2}}

I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }

I=0.0987 kgm^{2}

For Part (b) Angular Speed

From Kinetic Energy we get

KE=\frac{1}{2}IW^{2}

Pontential Energy

PE=mgd(1-Cosa)

KE=PE

\frac{1}{2}IW^{2}=mgd(1-Cosa)

W^{2}=\frac{2mgd(1-Cosa)}{I}

W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }

W=2.66\frac{rad}{s}

5 0
3 years ago
Directions: Summarize the main ideas of this lesson by answering the question below.
Allushta [10]

Answer:

Explanation:

Trypophobia is a fear or disgust of closely-packed holes. People who have it feel queasy when looking at surfaces that have small holes gathered close together.A person may start by imagining what they fear, then looking at pictures of the fear object, and then finally being near or even touching the source of their anxiety. In the case of trypophobia, a person with symptoms may start by simply closing his eyes and imagining something such as a honeycomb or seed poThey can help you find the root of the fear and manage your symptoms. Last medically reviewed on July 20, 2017 Medically reviewed by Timothy J. Legg, Ph.D., CRNP — Written by Annamarya Scaccia ...

8 0
3 years ago
A 770-kg boulder is raised from a quarry 106 m deep by a long uniform chain having a mass of 575 kg . This chain is of uniform s
Bas_tet [7]

Answer:

(a) a_{y}= 1.931m/s^{2}

(b) t=10.478s

Explanation:

For Part(a)

Apply ∑Fy=ma

T-Mg=Ma_{y} \\

Solve for a

a_{y}=\frac{T-Mg}{M}=\frac{2.80mg-Mg}{M} \\a_{y}=\frac{(2.80*575kg-1345kg)}{1345kg}*(9.80m/s^{2} )\\a_{y}= 1.931m/s^{2}

For Part(b)

Assume the maximum acceleration (maximum tension upwards is greater than the weight of the chain+boulder)

As the given data

a_{y}=1.931m/s^{2}\\  y-y_{o}=106m\\v_{oy}=0

Using equation of simple motion

y-y_{o}=v_{oy}t+(1/2)a_{y}t^{2} \\t=\sqrt{\frac{2(y-y_{o})}{a_{y}} }\\ t=\sqrt{\frac{2(106m)}{1.931m/s^{2} } }\\t=10.478s

4 0
4 years ago
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