Ang larong Latin at Sisiw ay_______________________

Two positive point charges repel each other with force 0.36 N when their separation is 1.5 m. What force do they exert on each o

egoroff_w [7]

The answer is: 0.81

I hope this helps :)

6 0

3 years ago

A projectile is fired from the origin (at y = 0 m) as shown in the diagram. The initial velocity

Viktor [21]

Answer:

-26 m/s.

Explanation:

Hello,

In this case, since the vertical initial velocity is 26 m/s and the vertical final velocity is 0 m/s at P, we compute the time to reach P:

$t=\frac{0m/s-26m/s}{-9.8m/s^2} =2.65s$

With which we compute the maximum height:

$y=26m/s*2.65s-\frac{1}{2}*9.8m/s^2*(2.65s)^2 \\\\y=34.5m$

Therefore, the final velocity until the floor, assuming P as the starting point (Voy=0m/s), turns out:

$v_f=\sqrt{0m/s-(-9.8m/s^2)*2*34.5m}\\ \\v_f=-26m/s$

Which is clearly negative since it the projectile is moving downwards the starting point.

Regards.

3 0

4 years ago

Which of the following is not a component of fitness? A. Cardiovascular fitness B. Muscular endurance C. Pulmonary endurance D.

Ne4ueva [31]

Answer:

C. Pulmonary endurance

Explanation:

I'm pretty sure it's "C" because cardiovascular and pulmonary endurance are the same thing and usually you'd hear cardiovascular more than pulmonary.

Sorry if I'm wrong!

7 0

3 years ago

If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in

motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

$F_{initial} = \frac{kq_1q_2}{r^2}$

Here,

k = Coulomb's constant

$q_{1,2}$ = Charge at each object

r = Distance between them

As the distance is doubled so,

$F_{final} = \frac{kq_1q_2}{( 2r )^2}$

$F_{final} = \frac{ kq_1q_2}{ 4r^2}$

$F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}$

$F_{final} = \frac{1}{4} F_{initial}$

$\frac{F_{final}}{ F_{initial}} = \frac{1}{4}$

Therefore the factor is 1/4

6 0

3 years ago

Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an

diamong [38]

Answer:

(a) Acceleration = 0.1063 m/s^2 (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration = 0.1063 m/s^2 (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0

4 years ago

Ang larong Latin at Sisiw ay________________________.