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quester [9]
1 year ago
6

(a) that the acceleration (A) will have no components that are identically zero. (B) may have some components that are identical

ly zero. (C) will have only a z component that is identically zero. (D) will have an identically zero z component, and maybe an identically zero component in the x or y directio
Physics
1 answer:
umka21 [38]1 year ago
4 0

Answer:

D.

Explanation:

acceleration in the z component = velocity in z component / time. The question states velocity in z component is zero. Therefore, acceleration in z component = 0/time = 0.

Now the question does not specify whether the other velocity vectors (x or y) is zero or not. So its D.

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That's "displacement". It only depends on the beginning and ending locations, and doesn't care about the route between them.
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3 years ago
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What does strong language mean ?
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Answer:

noun

the style of a piece of writing or speech.

"he explained the procedure in simple, everyday language"

coarse or offensive language.

noun: strong language

"strong language"

Explanation:

comment how it helps

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3 years ago
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25. Explain why the speed of sound is faster in solids than in gases. Include two other factors Chapter 17 says the speed of sou
aksik [14]
In solids, particles or atom are very closely arranged compared to gasses. When these particles are arranged in such proximity, vibrations from sound are very easily transmitted from one particle to another in the solid. Hence, the sound vibrations can travel through the solid medium more quickly than through a gas medium.
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7 0
3 years ago
Four blocks of weights are required using which any body whose weight is between 1kg and 40 kg can be weighed. Find the four wei
Vikentia [17]

Answer:

The weights are 1 kg, 3kg, 9kg and 27kg.

Explanation:

The weights are 1 kg, 3kg, 9kg and 27kg.

1+3+9+27= 40

27+9+3= 39

27+9+3-1=38

27+9+1=37

27+9=36

27+9-1=35

27+9+1-3=34

27+9-3=33

27+9-3-1=32

27+3+1=31

27+3=30

27+3-1=29

27+1=28

27

27-1=26

27+1-3=25

27-3=24

27-3-1=23

27+3+1-9=22

27+3-9=21

27+3-9-1=20

Like this all the weights from 1 to 40 kg can be made using 1,3,9 and 27 kg.

6 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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