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Yakvenalex [24]

2 years ago

9

Describe the way that distance is changing in relation to tiMe

1 answer:

AlekseyPX2 years ago

8 0

Answer:

DistaNcE chanGes iN relAtiOn tO TiMe Bc ThE fuRthEr yOu Go ThE mOrE TiMe goeS bY

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CAN SOMEONE PLEASE HELP ME

Anna71 [15]

Answer:

No, it is not fair because they could overdo it and get injured and others may not have it.

3 0

3 years ago

"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips an

tekilochka [14]

To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

$F=qE$

Here

q= Charge

E = Electric Force

$F=(1.6*10^{-19}C)(2750N/C)$

$F = 4.4*10^{-16}N$

PART B) Rearrange the expression F=ma for the acceleration

$a = \frac{F}{m}$

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

$a = \frac{4.4*10^{-16}N}{1.67*10^{-27}kg}$

$a = 2.635*10^{11}m/s^2$

PART C) Acceleration can be described as the speed change in an instant of time,

$a = \frac{v_f-v_i}{t}$

There is not $v_i$ then

$a = \frac{v_f}{t}$

Rearranging to find the velocity,

$v_f = at$

$v_f = (2.635*10^{11})(1.4*10^{-6})$

$v_f = 3.689*10^{5}m/s$

7 0

3 years ago

0.054x2.33x90............

guapka [62]

<h2>0.054×2.33×90</h2><h3>=0.11582×90</h3><h3>=11.3238</h3>

please mark this answer as brainlist

5 0

3 years ago

2. Which process is used by plants to cycle water from the ground, through the roots, and out the leaves as water vapor?

Salsk061 [2.6K]

Answer:

The answer is transpiration

5 0

3 years ago

Read 2 more answers

How fast, in meters per second, does an observer need to approach a stationary sound source in order to observe a 7.1 % increase

Nutka1998 [239]

Answer:

v0 = 24.42 m/s (Approx)

Explanation:

Given:

Increase in frequency = 7.1% =

Computation:

Assume n = 100%

n1 = [(v+v0)/(v+v1)]n

[100 + 7.1] = [(344+v0)/(344+0)]100

107.1 = [(344+v0)/(344)]100

v0 = 24.42 m/s (Approx)

5 0

4 years ago