Which of the following is not a function of PACs?

One of the checks that you could do for problem 1) would be to check the output resistance of the Wheatstone bridge to make sure

Ray Of Light [21]

Answer:

Explanation:

Let the four resistances of th wheat stone bridge is

P, Q, R and S and the value of each is 350 ohm.

Here, P and Q are in series.

R' = P + Q = 350 + 350 = 700 ohm

Then R and S are in series

R' = R + S = 350 + 350 = 700 ohm

Now R' and R'' are in parallel.

So, the equivalent resistance is

Req = R' x R'' / ( R' + R'')

Req = 700 / 2 = 350 ohm

Thus, the reading of ohmmeter is 350 ohm.

6 0

3 years ago

Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same r

Naily [24]

Answer:

The value of the centripetal forces are same.

Explanation:

Given:

The masses of the cars are same. The radii of the banked paths are same. The weight of an object on the moon is about one sixth of its weight on earth.

The expression for centripetal force is given by,

$F_{c} = \dfrac{mv^{2}}{r}$

where, $m$ is the mass of the object, $v$ is the velocity of the object and $r$ is the radius of the path.

The value of the centripetal force depends on the mass of the object, not on its weight.

As both on moon and earth the velocity of the cars and the radii of the paths are same, so the centripetal forces are the same.

3 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

4 years ago

The Cassegrain design provides more compact (shorter) telescopes. Why? (Examine figures 2.4.2 and 2.4.3). The shorter design is

Ipatiy [6.2K]

Answer:

Because the light reflects multiple times until it gets to the Cassegrain focus.

Explanation:

The Cassegrain design can be seen in a reflecting telescope. In this type of design the light is collected by a concave mirror, and then intercepted by a secondary convex mirror, and sends it down to a central opening in the primary mirror (concave mirror), in which a detector is placed (Cassegrain focus)

Since, the light is reflected many times due to Cassegrain design, that leads to shorter telescopes.

5 0

3 years ago

Based on observations, the speed of a jogger can be approximated by the relation v 5 7.5(1 2 0.04x) 0.3, where v and x are expre

castortr0y [4]

Answer:

solution:

to find the speed of a jogger use the following relation:

V

=

d

x

/d

t

=

7.5

×m

i

/

h

r

...........................(

1

)

in Above equation in x and t. Separating the variables and integrating,

∫

d

x

/7.5

×=

∫

d

t

+

C

or

−

4.7619

=

t

+

C

Here C =constant of integration.

x

=

0 at t

=

0

, we get: C

=

−

4.7619

now we have the relation to find the position and time for the jogger as:

−

4.7619 =

t

−

4.7619

.

(

2

)

Here

x is measured in miles and t in hours.

(a) To find the distance the jogger has run in 1 hr, we set t=1 in equation (2),

to get:

= −

4.7619

=

1

−

4.7619

= −

3.7619

or x

=

7.15

m

i

l

e

s

(b) To find the jogger's acceleration in m

i

l

/

differentiate

equation (1) with respect to time.

we have to eliminate x from the equation (1) using equation (2).

Eliminating x we get:

v

=

7.5×

Now differentiating above equation w.r.t time we get:

a

=

d

v/

d

t

=

−

0.675

/

At

t

=

0

the joggers acceleration is :

a

=

−

0.675

m

i

l

/

=

−

4.34

×

f

t

/

(c) required time for the jogger to run 6 miles is obtained by setting

x

=

6 in equation (2). We get:

−

4.7619

(

1

−

(

0.04

×

6 )

)^

7

/

10=

t

−

4.7619

or

t

=

0.832

h

r

s

6 0

3 years ago