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Ad libitum [116K]
2 years ago
8

A car goes from 0 to 100 km/hr in 10 seconds. What is its acceleration

Physics
1 answer:
makvit [3.9K]2 years ago
6 0

Answer:

a= 2.7 m/s^2

Explanation:

acceleration: a

speed: V

Vf = final speed

Vi= initial speed (initial = beginning)

100 km/hour --> m/s

divide the speed value by 3.6

100/3.6= 27.8 m/s

a = \frac{final speed - initialspeed}{time}

a= \frac{Vf - Vi}{t} \\

a= \frac{27-0}{10}

a= \frac{27}{10}

a= 2.7 m/s^2

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What were the two different alleles for height in the pea plants that Mendel studied?
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You might want to go to this website,       http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P
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3 years ago
A boy throws a stone straight upward with an initial
Dominik [7]

Answer:

14.8m

Explanation:

Given parameters:

Initial speed  = 17m/s

Unknown:

Maximum height  = ?

Solution:

At the maximum height, the final speed will be 0m/s;

 We use of the kinematics equation to solve this problem.

     V²   = U²   - 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

         0²   = 17²  -  (2  x 9.8 x h )

          0 = 289  - (9.6h)

         -289  = -19.6h

            h  = 14.8m

5 0
2 years ago
A small flashlight bulb provides a resistance of 3 Ω to the 300 mA current that runs through it. Determine the voltage of the ba
Contact [7]

V=IR

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3 0
3 years ago
An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runner’s average speed? (Round your answer to the ne
LUCKY_DIMON [66]
The answer that is got 8.7 . I got that because if you divide 200 by 23  you get <span>8.69565217391 and if you round that you get 8.7</span>
5 0
3 years ago
Read 2 more answers
A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t
Jobisdone [24]

Answer:

(a) E=0  :   0 cm from the center of the sphere

(b) E= 227.8*10³ N/C   :    10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C    :    40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C  :    59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

E=\frac{K*Q}{r^{2} } : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

E=k*\frac{Q}{a^{3} } *r :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q:  Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at  0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere  

r>a , We apply the Formula (1) :

E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }

E= 411.84 * 10³ N/C

4 0
3 years ago
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