When two ions with the same charge come together, their electric fields will

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straig

kupik [55]

Answer:

Explanation:

V = Deltax/Deltat

V = 15.0 m/s

Displacement:

(a) Vf = Vi + adeltat

Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s

Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m

(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s

Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m

(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s

Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m

(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s = 19.6m/s

Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m

4 0

4 years ago

Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on

nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂ -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C (OR) q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC (OR) q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

8 0

3 years ago

Doug, who runs track for his high school, was challenged to a race by his younger brother, matt. Matt started running first, and

eduard

The speed of Matt is 10 mph.

Doug runs 2 miles an hour faster than Matt, so let Matt’s speed equal x miles per hour. Then Doug’s speed equals x + 2 miles per hour. Each lap is one-quarter of a mile, so Doug runs 1.5 miles in the time it takes Matt to run 1.25 miles.

Rate of Matt is x

Rate of Dough is (x + 2)

Time taken by Matt is 1.25/x

Time taken by Dough is 1.25/(x + 2)

Distance covered by Matt is 1.25

Distance covered by Dough is 1.5

Dough and Matt took the same amount of time from the time Doug started, so make an equation by setting the two times in the chart equal to each other, and then solve for x:

$\frac{1.5}{(x + 2)}$ = $\frac{1.25}{x}$

1.5x = 1.25(x + 2)

1.5x = 1.25x + 2.5

0.25x = 2.5

x = 10

So Matt ran at 10 miles per hour.

To know more about time, speed and distance, visit: brainly.com/question/26046491

#SPJ4

7 0

1 year ago

Your aunt just turned 50 and now needs glasses to read.

alexira [117]

Answer:

(a) hypermetropia

(b) convex lens

(c) 133.33 cm

(d) - 21.05 cm

Explanation:

(a) As she is old age, so she is suffering from hypermetropia.

(b) It is the defect due to which a person is not able to see the nearby objects clearly, so it is cured by convex lens of suitable focal length.

(c) Power, P = + 0.75 D

Focal length is the reciprocal of power of lens.

f = 1/ P = 1/0.75

f = 133.33 cm

(d) v = -25 cm, f = 133.3 cm

use lens equation

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{133.3}=\frac{-1}{25}-\frac{1}{u}$

$\frac{-1}{u}=\frac{133.33+25}{133.33\times25}$

u = - 21.05 cm

8 0

3 years ago

A force of 4400 N is exerted on a piston that has an area of 0.020 m.

asambeis [7]

Use Pascal's Law.

F1/A1 = F2/A2

F1 = 4400N

A1 = 0.020m^2

F2 = X

A2 = 0.040m^2

4400/0.020 = X/0.040

Cross multiply:

(4400 * 0.040) = 0.020X

176 = 0.020x

Divide both sides by 0.020

X = 176 / 0.020

X = 8800

The force is 8800 N

3 0

3 years ago