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lozanna [386]
3 years ago
11

I need help fast please help

Physics
1 answer:
Irina18 [472]3 years ago
5 0
A and C Im pretty sure :)
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The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =
luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = \frac{\textup{Distance}}{\textup{Velocity}}

or

Time = \frac{\textup{0.016}}{\textup{23}}

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = \frac{q\times8.20\times10^4}{1\times10^{-11}}

Now from the Newton's equation of motion

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

where,  

d is the distance

u is the initial speed  

a is the acceleration

t is the time

or

0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2

or

q = 9.98 × 10⁻⁹ C

4 0
3 years ago
When a cannon is fired, how does the size of the force of the cannon on the cannonball compared with the force of the cannonball
kati45 [8]
The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
3 0
3 years ago
A typical adult can deliver about 12.5 N·m of torque when attempting to open a twist-off cap on a bottle. Assume that bottle cap
Nikitich [7]
Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh
3 0
3 years ago
A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___
34kurt

Answer:

The final displacement from the origin is <u>1.6</u> km to the <u>NE</u>

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8·\hat j, in vector form

2) 1.0 km East = 1.0·\hat i, in vector form

3) 1.6 km South = -1.6·\hat j, in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

Final displacement, = 1.0·\hat i + 1.2·\hat j

Where;

Δx = The displacement in the x-direction = 1.0·\hat i

Δy = The displacement in the y-direction = 1.2·\hat j

The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0
3 years ago
Please help! Will give brainliest. 10 points. Show work!
Natasha_Volkova [10]

Answer:

421.83 m.

Explanation:

The following data were obtained from the question:

Height (h) = 396.9 m

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

First, we shall determine the time taken for the ball to get to the ground.

This can be calculated by doing the following:

t = √(2h/g)

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 396.9 m

Time (t) =.?

t = √(2h/g)

t = √(2 x 396.9 / 9.8)

t = √81

t = 9 secs.

Therefore, it took 9 secs fir the ball to get to the ground.

Finally, we shall determine the horizontal distance travelled by the ball as illustrated below:

Time (t) = 9 secs.

Initial velocity (u) = 46.87 m/s

Horizontal distance (s) =...?

s = ut

s = 46.87 x 9

s = 421.83 m

Therefore, the horizontal distance travelled by the ball is 421.83 m

5 0
3 years ago
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