The pressure in a container will stay the same if you —

A box is being pulled to the right. What is the magnitude of the Kinect frictional force?

Anna35 [415]

The answer to this question is A - 25 N

3 0

3 years ago

Read 2 more answers

Select the correct answer.

Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

A.

We’ll see the bell move, but we won’t hear it ring.

B.

We won’t see the bell move, but we’ll hear it ring.

C.

We’ll see the bell move and hear it ring.

D.

We won’t see the bell move or hear it ring.

E.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

A.

We'll see the bell move, but we won’t hear it ring.

5 0

3 years ago

Read 2 more answers

While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.

lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0

3 years ago

How to ask a person out in a different way? LIke insted of do you want to go? Like i like this guy and i gave him a ig teddy bea

Greeley [361]

You should come off like would you like to hang out sometime

6 0

3 years ago

Read 2 more answers

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal

galben [10]

Hello

1) First of all, since we know the radius of the wire ( $r=1.2~mm=0.0012~m$ ), we can calculate its cross-sectional area
$A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2$

2) Then, we can calculate the current density J inside the wire. Since we know the current, $I=3~A$ , and the area calculated at the previous step, we have
$J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2$

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity $\sigma=3.6\cdot10^7~ \frac{A}{Vm}$ of the aluminium, the electric field is given by
$E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m$

4 0

3 years ago

The pressure in a container will stay the same if you ——?