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vlada-n [284]
3 years ago
11

1. According to paragraph 3 in the text, MOST of the electromagnetic waves from

Physics
1 answer:
Fed [463]3 years ago
5 0

Answer:

Most of the EM waves from the sun that reach Earth are infrared waves, visible light, and UV radiation.

Explanation:

I hope this helps! Have a good day!

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A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th
Pavel [41]

Answer:

The block's mass should be 3m

Explanation:

Given:

Cart with mass m

From the conservation of energy before mass is added,

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

Where A = amplitude of spring mass system, k = spring constant

  A = v\sqrt{\frac{m}{k} }

Now new mass M is added to the system,

   \frac{1}{2} (m +M ) v^{2}  = \frac{1}{2}  k A^{2}

  A = v \sqrt{\frac{m +M }{k} }

Here, given in question frequency is reduced to half so we can write,

   f' = \frac{f}{2}

Where f = frequency of system before mass is added, f' = frequency of system after mass is added.

        \omega ' = \frac{\omega}{2}

\sqrt{\frac{k}{m +M} }  = \frac{\sqrt{\frac{k}{m} } }{2}

   \frac{k}{m +M } = \frac{k}{4m}

   M = 3m

Therefore, the block's mass should be 3m

8 0
3 years ago
A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3
I am Lyosha [343]

Answer:

F_{braking}=337299 pdl

Explanation:

Impulse-Momentum relation:

I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})

F_{total}=-F_{braking}+mgsin{\theta}

We solve the equations in order to find the braking force:

F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl

6 0
3 years ago
Which statements below are true?
Ilya [14]

The correct statements are that the speed decreases as the distance decreases and speed increases as the distance increases for the same time.

Answer:

Option A and Option B.

Explanation:

Speed is defined as the ratio of distance covered to the time taken to cover that distance. So Speed = Distance/Time. In other words, we can also state that speed is directly proportional to the distance for a constant time. Thus, the speed will be decreasing as there is decrease in distance for the same time. As well as there will be increase in speed as the distance increases for the same time. So option A and option B are the true options. So if there is decrease in the distance due to direct proportionality the speed will also be decreasing. Similarly, if the distance increases, the speed will also be increasing.

7 0
3 years ago
Two spherical objects have masses of 6.2 x 105 kg and 13 x 103 kg. The gravitational attraction between them is 130 N. How far a
Neporo4naja [7]

Answer:

r = 6.4 cm

Explanation:

F = GMm/r²

r = √(GMm/F)

r = √((6.674e-11)(6.2e5)(13e3)/130)

r = 0.06432... m

Those are some high density materials!

8 0
3 years ago
A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo
statuscvo [17]
The Young modulus E is given by:
E= \frac{F L_0}{A \Delta L}
where 
F is the force applied
A is the cross-sectional area perpendicular to the force applied
L_0 is the initial length of the object
\Delta L is the increase (or decrease) in length of the object.

In our problem, L_0 = 3.0 m is the initial length of the column, E=3.0 \cdot 10^{10}N/m^2 is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m
and the cross-sectional area is
A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2
The force applied to the column is the weight of the load:
W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N

Now we have everything to calculate the compression of the column:
\Delta L =  \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m
So, the column compresses by 1.83 millimeters.
3 0
3 years ago
Read 2 more answers
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