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mr Goodwill [35]
3 years ago
15

I NEED HELP ASAP!!!

Physics
2 answers:
Tamiku [17]3 years ago
7 0
I think option C is correct in my opinion
Savatey [412]3 years ago
3 0
I think option C is correct..hope it helps
You might be interested in
An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi
Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

7 0
3 years ago
Radon-222 ( 222/86 Rn) is a radioactive gas with a half-life of 3.82 days. A gas sample contains 4.1 e 8 radon atoms initially.
kow [346]

Answer :

(a) The number of radon atoms will remain after 12 days is, 4.67\times 10^7

(b) The number of radon nuclei have decayed by this time will be, 3.6\times 10^8

Explanation :

<u>For part (a) :</u>

Half-life = 3.82 days

First we have to calculate the rate constant, we use the formula :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{3.82\text{ days}}

k=1.81\times 10^{-1}\text{ days}^{-1}

Now we have to calculate the number of radon atoms will remain after 12 days.

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = 1.81\times 10^{-1}\text{ days}^{-1}

t = time passed by the sample  = 12 days

a = initially number of radon atoms  = 4.1\times 10^8

a - x = number of radon atoms left = ?

Now put all the given values in above equation, we get

12=\frac{2.303}{1.81\times 10^{-1}}\log\frac{4.1\times 10^8}{a-x}

a-x=4.67\times 10^7

Thus, the number of radon atoms will remain after 12 days is, 4.67\times 10^7

<u>For part (b) :</u>

Now we have to calculate the number of radon nuclei will have decayed by this time.

The number of radon nuclei have decayed = Initial number of radon atoms - Number of radon atoms left

The number of radon nuclei have decayed = (4.1\times 10^8)-(4.67\times 10^7)

The number of radon nuclei have decayed = 3.6\times 10^8

Thus, the number of radon nuclei have decayed by this time will be, 3.6\times 10^8

5 0
3 years ago
Which of the following is not a main function of the human's body nervous system?
Kipish [7]

Answer:

I can't see option dear

Explanation:

6 0
3 years ago
Could you please solve it with shiwing the full work
tia_tia [17]

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx  :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5  = 49 m/s : variable velocity in y

v=\sqrt{v_{x} ^{2} +v_{y} ^{2} }

v=\sqrt{ 638.6^{2} +49 ^{2} }

V= 640.48 m/s : total velocity in t= 5s

2. v_{ox} =v_{o} cos33.2=20.9*cos33,2= 17.49 m/s

v_{oy}=v_{o}*sin33,2 =20.9*sin33,2=11.44 m/s

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4.  a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0  , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

5 0
3 years ago
you drive in a straight line at 20.0m/s for 10.0 min, then 30.0m/s for another 10.0 min. Find the average speed.
Juliette [100K]

Answer is 25 m/s.

<em>Explanation;</em>

<em>Step 1 : Convert the units into same unit.</em>

Here speeds are given in m/s while the time is in minutes. So, let's convert minutes into seconds.

1 min = 60 s

Then,

   10 minutes = 10 x 60 s = 600 s

<em>Step 2 :  Find the total distance that you have travelled.</em>

To find that you need to first find distances separately.

We know the formula, s = vt, where s is the distance, v is the speed or velocity and t is the time.

  • First you have travelled 10 minutes (600 s) at 20.0 m/s.

        Then distance, s = 20.0 m/s x 600 s = 12 000 m

  • Secondly you have travelled again 10 minutes at 30.0 m/s

        Then distance = 30.0 m/s x 600 s = 18 000 m

Then, the total distance = 12 000 m + 18 000 m = 30 000 m

<em>Step 3 : Find the total time taken to travel.</em>

Total time  = 600 s + 600 s = 1200 s

<em>Step 4 : Calculate the average speed.</em>

Average speed = total distance / total time

                          = 30 000 m / 1200 s

                          = 25 m/s


   

6 0
3 years ago
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