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galina1969 [7]
4 years ago
13

Sound waves cannot travel in a vacuum because

Physics
1 answer:
blsea [12.9K]4 years ago
4 0
A. Nothing for them to travle through,
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14. A block rests on a frictionless table on Earth. After a 20-N horizontal force is applied to the block, it accelerates at 3.9
joja [24]
<span>14. Let's find mass. We know that F = m*a, so m = F/a.m = 20/3,9 m/s^2 = 5,12 kg;We already know the mass, so acceleration will be a = F/m;a = 10/5,12 = 2,0 m/s^2 (approximately);Answer: E.
15. According to the picture given above, let's define all components:These are refer to y: 985N sin(31)= 507;788N cos(32)= 668;411N sin(53)=-328;
And these are refer to x: 985N cos(31)= 844 788N sin(32)= -417 411N cos(53)= -247
And finally let's calculate tan:  tan^-1 (507/844)= 30,98; tan^-1 (668/-417)= -58,02; tan^-1 (-328/-247)= 53,01
According to these calculations we've got: Sum Fx = 179,4 and Sum Fy = 847. Then let's count magnitude:Fsum = rad 179^2 + 847^2 = 865
Then we've got [tex]cos^-1 (179/865) = 77.7 (approximately).
So the most approximate answer is C. 866 N at 78.1° counterclockwise to the x-axis.
16. I think that this question is incomplete because there wasn't mentioned in what end of the chain the tension should be calculated. Anyway I'll help you with both bottom and top ends. We will use the basic formula W=m*g; W(bottom) = 175*9.8 = 1715 N; W(top) = (175+12)*9.8 = 1833 N; So you should choose between B. 1830 N or D. 1720 N. But I think the most possible answer is B.
17. I am definitely sure that A diagram generates the most tension in one chain. So the answer is C. The box is held by 1 chain and have all its weight.
18. I think that each planet would move in a straight line at constant speed, because there will be a zero gravity condition and there won't any impact on the planets.
19.  According to the Work-Energy Theorem we have:1/2*1200*2^2 = 2400 (J);Then let's count the average force according to the formula: Work =  F*D where D is displacement.F = 2400/ 0.15 = 16000 (N)
So the answer is B. 1.6 × 10^4 N
20. If my memory serves me well, magnetism is the force that can act through empty space. It's one aspect of the combined electromagnetic force. So the answer is A.
21. According to the illustration given above, I think we should use formula F = m*g; Let's count m. m = 5 kg - 0,6 kg = 4,4 kg; Then we have everything to count force:F = 4,4 * 9,8 = 43,1 N. So the most approximate answer is D. 43 N.
22. I am definitely sure that the answer is C. on Earth at sea level. Because weight has the formula W=mg. And on the earth surface the magnitude of g is higher that everywhere so the greatest weight is on Earth at sea level.
23. We have everything to calculate the acceleration. According to Newton's second law, the formula is a = F/m;a = 20/40 = 0,5 m/s^2; So the answer is A.
24. According to the definition of action reaction forces, the answer should be: A. The forces are opposite in direction, with the force on the ball much stronger in magnitude.
25. I am pretty sure that we should use Newton’s second law of motion F = m*a. First we should find mass, using formula W=m*g => m = W/g => m = 2400/9.8= 245 kg. Then we can find F.F = m*a;F = 245*12 = 2940 N; Answer: B
</span>
7 0
3 years ago
Read 2 more answers
What is in between the nucleus and the electrons in an atom?
hoa [83]

Answer:

D. Empty Space

Explanation:

4 0
3 years ago
A car with an initial speed of 31.4 km/h accelerates at a uniform rate of 1.2 m/s2 for 1.3 s. What is the displacement of the ca
Kaylis [27]

Answer:

21.5 m

Explanation:

A car has an initial speed of 31.4 km/hr

Convert to m/s

= 31.4 × 1000/3600

= 31,400/3600

= 8.722 m/s

Acceleration = 1.2 m/s^2

Time= 1.3 seconds.

Therefore the displacement can be calculated as follows

S= 8.722 × 1.3 + 1/2 × 1.2 × 1.3^2

= 11.34 + 1/2 × 20.28

= 11.34 + 10.14

= 21.5 m

5 0
3 years ago
Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
zysi [14]

Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
3 years ago
What is the Sun's right ascension on the vernal equinox?
Kay [80]
The suns right ascension on the vernal equinox is (a) 0 hours



3 0
3 years ago
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