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vlada-n [284]
2 years ago
8

A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the

block and the surface has magnitude 8 N. If the direction of the applied force is reversed which of the following is true?
A. the magnitude of the velocity of the block will remain constant.
B. the magnitude of the acceleration of the block decreases
C. The magnitude of the net force on the block increases as compared to before.
D. the magnitude of the net force on the block will remain the same as before​
Physics
1 answer:
kompoz [17]2 years ago
3 0

' A ' and ' D ' are both correct statements.

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What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w
Tatiana [17]

Answer:

a. \ f_1=7.9057Hz\\\\b. \ f_2=15.8114Hz\\\\c. \ f_3=23.7171Hz

Explanation:

a. The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu} #where t=250N*10=2500N, \mu=0.1kg

#substitute for actual values for the lowest frequency.

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{1}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=7.9057Hz  #n=1, lowest frequency

Hence, the lowest frequency for standing waves is 7.9057Hz

b.The wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}           #where t=250N*10=2500N,\mu=0.1kg

#The second lowest frequency happens at n=2:

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{2}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=15.8114Hz

Hence, the second lowest frequency is 15.8114Hz

c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.

Frequency is given by the equation:

f=\frac{nv}{2L}\\\\=\frac{n}{2L}\sqrt{\frac{t}{\mu}                    #where t=250N*10=2500N,\mu=0.1kg

The third lowest frequency happens at n=3

F=\frac{n}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=\frac{3}{2\times 10m}\times \sqrt{2500N/0.1kg}\\=23.7171Hz

Hence, the third lowest frequency is 23.7171Hz

6 0
3 years ago
What is the force on a 52 coulomb charge in a field of 5 N/C
slega [8]
Https://courses.physics.illinois.edu/phys102/sp2013/lectures/lecture2.pdf



Check out upload.
Download pdf
8 0
3 years ago
Jim is driving a 2268-kg pickup truck at 22 m/s and releases his foot from the accelerator pedal. The car eventually stops due t
shutvik [7]

Answer:

610 meters.

Explanation:

Because Jim released the accelerator, the truck started to slow down, so the friction force will eventually stop the truck.

the kinetic energy of the truck just after Jim released the pedal is:

E_k=\frac{1}{2}*m*v^2\\E_k=\frac{1}{2}*2268*(22)^2=548856J

The work done by the friction force is given by:

W_f=F_s*d\\\\d=\frac{548856J}{900N}\\\\d=610m

6 0
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Air in the atmosphere is heated by the ground this warm air then rises and cooler air falls this is an example of what type of p
Alona [7]
The answer for both is ‘B’
8 0
2 years ago
Electromagnetic waves are ........... by shiny surfaces.
sleet_krkn [62]

Answer:

Electromagnetic waves are reflected

Explanation:

Reflection of light (and other forms of electromagnetic radiation) occurs when the waves encounter a surface or other boundary that does not absorb the energy of the radiation and bounces the waves away from the surface. ... This concept is often termed the Law of Reflection.

4 0
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