Easy one here pls help thanks so much

Which tagline from a car advertisement has the BEST connotative appeal to the parents of a new teenage driver?

Nonamiya [84]

A. Advanced new safety features that get everyone home at night.

6 0

3 years ago

Please help me !!!!!!

s2008m [1.1K]

Answer:

A

Explanation:

A disadvantage is that sexual reproduction takes longer than asexual reproduction. A mate must be found, the egg must be fertilised by sperm, and then the offspring develop. ... This means that some individuals would survive and be able to reproduce and generate more offspring.

5 0

3 years ago

A water line with an internal radius of 5.29 x 10-3 m is connected to a shower head that has 15 holes. The speed of the water in

fiasKO [112]

Answer:

(a) 3.44 x 10^-3 m^3/s

(b) 8.4 m/s

Explanation:

area of water line, A = 5.29 x 10^-3 m

number of holes, N = 15

Speed of water in line, V = 0.651 m/s

(a) Volume flow rate is given by

V = area of water line x speed of water in water line

V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s

(b) area of one hole, a = 4.13 x 10^-4 m

Let v be the velocity of water in each hole

According to the equation of continuity

A x V = a x v

5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v

v = 8.4 m/s

5 0

3 years ago

Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0

lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0

3 years ago

If the initial velocity of a ball is sent straight upward at 10.5m/s from the ground what will its final velocity be when it hit

Shalnov [3]

Answer: -10.08 m/s

Explanation:

Here we only need to analyze the vertical problem.

When the ball is in the air, the only force acting on it will be the gravitational force, this means that the acceleration of the ball, is equal to the gravitational acceleration, then:

a(t) = -9.8m/s^2

Where the negative sign is because gravity pulls the ball down.

To get the velocity equation we need to integrate over time, we get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial vertical velocity, here it is v0 = 10.5 m/s

Then the velocity equation is:

v(t) = (-9.8m/s^2)*t + 10.5 m/s

To get the position equation, we need to integrate again over time, we get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t + p0

Where p0 is the initial position, we know that the ball is sent upward from the ground, so p0 = 0m

Then the position equation is:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

Now we need to find the value of t such that the position is equal to zero (this means that the ball hits the ground again).

Then we need to solve:

p(t) = 0 = (1/2)*(-9.8m/s^2)*t^2 + (10.5 m/s)*t

If we divide both sides by t, we get:

0 = (1/2)*(-9.8m/s^2)*t + (10.5 m/s)

Now we can solve it:

(1/2)*(9.8m/s^2)*t = 10.5 m/s

t = (10.5 m/s)/((1/2)*(9.8m/s^2)) = 2.14 s

This means that after 2.14 seconds, the ball will hit the ground again.

The velocity of the ball when it hits the ground is equal to:

v(2.14s) = (-9.8m/s^2)*2.14s + 10.5 m/s = -10.08 m/s

3 0

3 years ago