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myrzilka [38]
3 years ago
7

Why can't I get a new phone

Physics
2 answers:
Alenkasestr [34]3 years ago
3 0

Answer:

Idk I think we all ask the question also this question is for sure prob getting deleted just saying now not by me

Explanation:

kodGreya [7K]3 years ago
3 0

Answer:cause u can’t

Explanation:

U don’t have money to pay it

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Would u rather/content creator or a rap artist
BartSMP [9]

a content creator because if i was a rapper i probably wouldn't make good songs lol

7 0
3 years ago
A 0.683 kg mass moves in SHM at the end of a spring. It takes 1.41 s to move from the position with the spring fully extended to
dsp73

Answer:

Spring constant of the spring will be equal to 9.255 N /m          

Explanation:

We have given mass m = 0.683 kg

Time taken to complete one oscillation is given T = 1.41 sec

We have to find the spring constant of the spring

From spring mass system time period is equal to T=2\pi \sqrt{\frac{m}{K}}, here m is mass and K is spring constant

So 1.41=2\times 3.14\sqrt{\frac{0.683}{K}}

0.2245=\sqrt{\frac{0.683}{K}}

Squaring both side

0.0504=\frac{0.4664}{K}

K=9.255N/m

So spring constant of the spring will be equal to 9.255 N /m

7 0
3 years ago
What is the relationship between amplitude and frequency of a wave?.
kondaur [170]

Answer:

The relationship between the wave's amplitude and frequency is such that it is inversely proportional to the frequency. The amplitude decreases as the frequency increases. The amplitude increases as the frequency decreases. The higher the energy of a wave, the higher the amplitude. The lower the energy, the lower the amplitude. Energy has no effect on wavelength, speed, or frequency, only the amplitude.

Explanation:

5 0
2 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
What is the approximate wavelength of a light whose second-order dark band forms a diffraction angle of 15.0° when it passes thr
anzhelika [568]
414nm just took the test
7 0
3 years ago
Read 2 more answers
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