Write the dimensional formula of presor and gravitation constant

Technician A says that gears may be used to apply torque to other rotating parts. Technician B says that gears may be used to mu

Goryan [66]

Answer:

Both technicians are right.

Explanation:

Torque is defined as a rotational force that can be calculated by the formula T= F.d. Being F the force applied to a body to make it rotate and d the distance since the force is applied.

$T= F . d$

So technician A is saying that gear can apply a torque to another gear, and that is true because it’s applying a rotational force to the gear next to it, in simple terms -anytime I make something rotate, I'm generating a torque-. And for technician B, the torque also can vary as a function of the distance of the force applied (size of the gear), so, it can multiply the torque and change the rotation speed.

5 0

4 years ago

Whats the purpose of the metal strip under a car?

mezya [45]

Answer:

It ensures the doors seal properly when closed. This seal will fend off the elements, and prevent water from leaking inside the vehicle. It is further used on windows to seal and protect the windows.

Explanation:

3 0

3 years ago

Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,

MAVERICK [17]

Question is missing. Found on internet the complete text of the problem:

"Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
 $E= \frac{\sigma}{2 \epsilon _0}$
where $\sigma$ is the charge density and $\epsilon _0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are $\sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2$ and $\sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2$ .

Sheet a gives an electric field of
$E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m$
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
$E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m$
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
$E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m$
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
$E=E_1+E_2=-11.98\cdot 10^5 V/m$
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
$E=E_1+E_2=11.98 \cdot 10^5 V/m$
towards left.

7 0

4 years ago

What is the displacement of the object from 12 seconds to 16 seconds

spayn [35]

Answer:

The displacement of the object is -8 m

Explanation:

Displacement

The displacement of a moving object can be calculated as the area under (or above) the graph of velocity vs time.

If the area is below the y-axis, then the displacement is negative. Otherwise is positive.

It's important to differentiate displacement from distance. Displacement takes into consideration the direction of the movement. Distance does not and it's always positive.

From the graph provided, we can see the velocity from t=12 s from t=16 s is negative, and the displacement will also be negative.

The displacement is calculated as the area of the triangle with base b=16-12= 4 seconds and height = -4 m/s, thus:

$\displaystyle D = \frac{4*(-4)}{2}=-\frac{16}{2}=-8\ m$

The displacement of the object is -8 m

8 0

3 years ago

A potential difference of 300 volts is applied to a 2.0-µf capacitor and an 8.0-µf capacitor connected in series. (a) What are t

zubka84 [21]

Answer:

a) Q1=Q2=480μC V1=240V V2=60V

b) Q1=96μC Q2=384μC V1=V2=48V

c) Q1=Q2=0C V1=V2=0V

Explanation:

Let C1 = 2μC and C2=8μC

For part (a) of this problem, we know that charge in a series circuit, is the same in C1 and C2. Having this in mind, we can calculate equivalent capacitance first:

$C=\frac{1}{\frac{1}{C1} +\frac{1}{C2} } = 1.6\mu C$

$Q_{T} = V_{T}*C_{T} = 480\mu C$

$V1 = \frac{Q1}{C1}=240V$

$V2 = \frac{Q2}{C2}=60V$

For part (b), the capacitors are in parallel now. In this condition, the voltage is the same for both capacitors:

$V1=V2$ So, $\frac{Q1}{C1} = \frac{Q2}{C2}$

Total charge is the same calculated for part (a), so:

$\frac{Qt - Q2}{C1} = \frac{Q2}{C2}$ Solving for Q2:

Q2 = 384μC Q1 = 96μC.

Therefore:

V1=V2=48V

For part (c), both capacitors would discharge, since their total voltage of 300V would by applied to a wire (R=0Ω). There would flow a huge amount of current for a short period of time, and capacitors would be completely discharged. Q1=Q2=0C V1=V2=0V

3 0

3 years ago

Write the dimensional formula ofpresor and gravitation constant​

Write the dimensional formula ofpresor and gravitation constant