What is the total voltage of the circuit? (must include unit

What is the maximum distance that a 60.-watt motor may vertically lift a 90.-newton?weight in 7.5 seconds?a. 2.3mb. 5.0mc. 140md

Naddik [55]

Answer:

b. 5.0 m

Explanation:

Power: Power can be defined as the rate at which work is done.

The S.I unit of power is watt (W).

P = W/t ............................................... Equation 1

Where p = power, W = work, t = time.

But

W = F×d..................................... Equation 2

Where F = force, d = maximum distance.

Substituting equation 2 into equation 1

P = F×d/t

making d the subject of the equation above,

d = Pt/F......................................................... Equation 3

Given: F = 90 Newton, P = 60 watt, t = 7.5 seconds.

Substituting these values into equation 3,

d = 60×7.5/90

d = 5 m

Thus the maximum distance is = 5 m

The right option is b. 5.0 m

5 0

3 years ago

All these are properties except

marin [14]

Amethyst... because it's silicon dioxide

3 0

3 years ago

In a loop-the-loop ride a car goes around a vertical, circular loop at a constant speed. The car has a mass m = 268 kg and moves

11111nata11111 [884]

Before we go through the questions, we need to calculate and determine some values first.

r = 11.5 m
m = 280 kg 
Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N

1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.

Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N 

2) What is the magnitude of the normal force on the car when it is at the side of the circle. 

Centripetal force = 7119.55 N 

3) What is the magnitude of the normal force on the car when it is at the top of the circle. 

Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N 

4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. 

√(gr) 

√(9.8 x 11.5) = 10.62 m/s

7 0

2 years ago

three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a

valentinak56 [21]

Answer:

$q_1=+0.375\ {10}^{-9}$

Explanation:

Electrostatic Forces

The force exerted between two point charges $q_1$ and $q_2$ separated a distance d is given by Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{d^2}$

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

$\displaystyle q_3=+5\ 10^{-9}\ c$

$\displaystyle q_2=-3\ 10^{-9}\ c$

The distance between $q_3$ and $q_2$ is

$\displaystyle d_2=4cm=0.04\ m$

The distance between $q_3$ and $q_1$ is

$\displaystyle d_1=2cm=0.02\ m$

We must find the value of $q_1$ such that

$\displaystyle |F_3|=0$

Applying Coulomb's formula for $q_1$ is

$\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}$

Now for $q_2$

$\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}$

If the total force on $q_3$ is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

$\displaystyle F_{13}=F_{23}$