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o-na [289]
3 years ago
8

Charge: A piece of plastic has a net charge of +2.00 μC. How many more protons than electrons does this piece of plastic have? (

e = 1.60 × 10-19 C)
Physics
1 answer:
snow_lady [41]3 years ago
6 0

Answer

given,

net charge = +2.00 μC

we know,

1 coulomb charge =  6.28 x 10¹⁸electrons

1 micro coulomb  charge =  6.28 x 10¹⁸ x 10⁻⁶ electron

                                         = 6.28 x 10¹² electrons

2.00 μC = 2 x 6.28 x 10¹² electrons

              = 1.256 x 10¹³ electrons

since net charge is positive.

The number of protons should be 1.256 x 10¹³ more than electrons.

hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.

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alexira [117]
<span>Na+/K+ ATPase (sodium pump) is considered to be an electrogenic pump because during every transport cycle, it transports 3 Na+ ions out of the cell and 2 K+ ions into the cell. This leads to the movement of one net positive charge out of the cell making this process electrogenic.</span>
3 0
3 years ago
2.) The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his/her he
Ratling [72]

Answer:

The minimum average speed the opponent must move so that he is in position to hit the ball is approximately 5.79 m/s

Explanation:

The given parameters of the ball are;

The initial speed of the ball = 15 m/s

The direction in which the ball is launched = 50° above the horizontal

The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

y = (u·sinθ)·t - 1/2·g·t²

When y = 2.1 m, we have;

2.1 = (15·sin(50°))·t - 1/2·9.8·t²

∴ 4.9·t² - (15·sin(50°))·t + 2.1 = 0

Solving with the aid of a graphing calculator function, we get;

t = 0.199776187257 s or t = 2.14525782198 s

Therefore, the ball is at 2.1 m above the start point on the other side of the court at t ≈ 2.145 seconds

The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

x = u × cos(50°) × t = 15 × cos(50°) × 2.145 ≈ 20.682 m

The horizontal distance the ball travels at t ≈ 2.145 seconds, x ≈ 20.682 m

Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

The minimum average speed the other player has to move with, v_s = d/t₂

∴ v_s = 10.682 m/(1.845 s) ≈ 5.78970189702 m/s ≈ 5.79 m/s

The minimum average speed the opponent must move so that he is in position to hit the ball, v_s ≈ 5.79 m/s.

5 0
3 years ago
What is the word that goes with the definition “ a disturbance in space that causes a force”
irga5000 [103]

Answer:

Gravity

Explanation:

Gravity is the force that causes a "disturbance" in space between planets.

8 0
3 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete
Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

a=\frac{v^2}{r}

where:

v is the speed

r is the radius

Now,

a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2

In g units:

a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g

7 0
3 years ago
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5
Leto [7]

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

\alpha = \frac{w2^{2}-w1^{2}  }{2*(\theta2 - \theta1)}

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec  /  7.67 x10^4 rad/sec²

t=1.097 sec

4 0
3 years ago
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