) Distinguish between scalar and vector<br>quantities<br>

s344n2d4d5 [400]

Answer:

b. The internal resistance must be much smaller than the other resistances in the circuit.

Explanation:

Ammeter is used to measure the current flowing through a circuit. It is connected in series configuration with the load. In such a scenario the resistance of the ammeter should be negligible so as to make sure that the voltage drop across the resistance of ammeter is zero and it shows the correct reading of the current in the circuit.

3 0

3 years ago

What two types of scientific knowledge can be expressed as mathematical equations? (Please don't tell me the answer could you pl

gtnhenbr [62]

An example would be 2 types of motion. It could be rectilinear or projectile motion. There are various equations for each type. Since you don't want me to tell you the answer, I could just express it in words. Then, it will be up to you to translate into mathematical equations.

For rectilinear motion, the distance traveled is equal to the initial velocity times the time, plus one-half of the acceleration times the square of the time. For projectile motion, the maximum distance is equal to the square of the initial velocity multiplied with the square of the sine of the launch angle, all over twice the gravity.

5 0

3 years ago

As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th

Lena [83]

Answer:

Pressure will be $6072449.952Pa$

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

$P=\frac{F}{A}=\frac{mg}{A}$

So force F = mg = 65×9.8 = 637 N

Area $A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2$

So pressure $p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa$

5 0

3 years ago

*please please help* Suppose you are going on a car trip with your family.

Strike441 [17]

Answer:

1. To determine the average speed for the first day of the trip, the total distance traveled would have to be acquired and then how long it took to arrive at the final destination, only including the time that was actually traveled and not any time that was accumulated by any rest stops. Once you have this information, you have to divide the distance over time and you have the average speed (mph).

2. To determine the instantaneous speed, you would just have to look at the speedometer, which tells you at what speed the car is traveling at that exact moment.

Explanation:

I took physics 121 and got the same question. This is my answer that i used and my teacher said it was right.

3 0

2 years ago

A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf

stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

$F - mg sin \theta = 0$

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta=30.0^{\circ}$ is the angle of the incline

Solving for F,

$F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N$

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

$P=Fv = (4655)(2.20)=10241 W$

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

$v=u+at$

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

$a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2$

So now the equation of the forces along the direction parallel to the incline is

$F - mg sin \theta = ma$

And solving for F, we find the maximum force applied by the motor:

$F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N$

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

$P=Fv=(4829)(2.20)=10624 W$

(c) $5.82\cdot 10^6 J$

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

$\Delta U = mg \Delta h$