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kenny6666 [7]
3 years ago
11

How much is required to accelerate a 5kg mass at 20 m/s2

Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0

use F = ma

F : force m : mass a : acceleration

so

f = 5kg * 20 m/s2 = 100 N

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i got the right answer for this question (pretty sure i guessed or just asked me teacher the answer?) but can someone explain ho
loris [4]
A guessing answer the best answer but you have had to subtract the answer by the equation that it was giving u
6 0
3 years ago
Read 2 more answers
THE LARGEST HORIZONTAL PLATES IS SEPARATED BY 4mm .The plate is at the potencial of -6V . What potencial should be applied to th
Serhud [2]

Answer:

V₂ = -22 V

Explanation:

Electric potential and field are related

         ΔV = - E d

where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates

 

In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m

          V₂- V₁ = - E d

          V₂ = - Ed + V₁

          V₂ = - 4000 0.004 + (- 6)

          V₂ = -16 - 6

          V₂ = -22 V

6 0
2 years ago
A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed
Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
A parallel-plate capacitor has an area of 4.59 cm2, and the plates are separated by 1.28 mm with air between them. it stores a c
nalin [4]
 <span>a) 
Capacitance = k x ε° x area / separation 
ε° = 8.854 10^-12 F/ m 
k = 2.4max 
average k = 0.78 / 1.27 * 2.4 +(1.27- 0.78) / 1.27 * 1 = 1.474 + 0.386 = 1.86 
(61.4 % separation k = 2.4 --- 38.6 % k = 1 air --- average k = 0.614 * 2.34 + 0.386 * 1 = 1.86 
area = 145 cm2 = 0.0145 m2 
separation = 1.27 cm 0.0127 m 

C = 1.86 * 8.854 10^-12 * 0.0145 / 0.0127 = 18.8 pF 
b) Q = C * V --- 18.8 * 83 = 1560.4 pC = 1.5604 nC 
c) E = V / d = 83 / 0.0127 = 6535.4 V/m </span>
7 0
2 years ago
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