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4 years ago

Why does an object hover over a large magnet

Physics

1 answer:

Anarel [89]4 years ago

7 0

Because the object hovering over the magnet is the same magnetic pole.

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A sinusoidal traveling wave is generated on a string by an oscillating source that completes 116 cycles per minute. What is the

slava [35]

Answer:

$\lambda =6.32\ cm$

Explanation:

given,

number of cycle complete (f) = 116 cycles per minute

wavelength observed at 11 m in 1.5 m.

$v = \dfrac{distance}{time}$

$v = \dfrac{11}{1.5}$

v = 7.33 m/s

$\lambda = \dfrac{v}{f}$

$\lambda = \dfrac{7.33}{116}$

$\lambda =0.0632\ m$

$\lambda =6.32\ cm$

The wavelength of the wave is equal to $\lambda =6.32\ cm$

6 0

4 years ago

Explain how you can cause light separated by a prism to combine

r-ruslan [8.4K]

When you shine a lite through a prism is reflects out light through all of the edges and causes light separation. Or just simply shine a laser through the edge of a sideways piece of glass.

I hope that this was helpful for you.

7 0

3 years ago

The graph below shows the relationship between speed and time for two objects, A and B. Compare with the acceleration of object

kolbaska11 [484]

Answer:

A) greater

Explanation:

acceleration is calculated by dividing velocity over time..so by calculating, you find acceleration of A is greater than that of B

5 0

4 years ago

Read 2 more answers

Which statement describes the distribution of charge in an atom? (1) A positively charged nucleus is surrounded by one or more n

pantera1 [17]

Answer: option B

Explanation: when a neutral atom loses an electron or gains a positive charge electron, it becomes a positive ion (positively charged) and when an neutral atom gains an electronic charge or losses a positive charge electron, it becomes a negative ion (negatively charged).

4 0

3 years ago

Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago