A horizontal line on a displacement-time graph means the object is

Tp-18 what should you do when operating a boat in large waves and high wind?

deff fn [24]

1) Try to head into the waves at some slight angle and the speed of the boat should be reduced.

2) In order to ride up and over the waves, the speed of the boat should be slow.

3) The less the speed of the boat, and the less strain will be put on the hull and superstructure.

5 0

3 years ago

A 70 kilogram hockey player skating east on an ice rink is hit by a 0.1 kilogram hockey puck moving toward the west. The puck ex

guajiro [1.7K]

Given:

70 Kilogram hockey

Hit by 0.1 kilogram

50 Newton

Description:

So in this case this bacially means that one object is experiencing a force then another object. So the answer to the question will be 50N.

Answer: 50N (3rd Law)

Hope this helps.

4 0

3 years ago

Can any juniors who go to Texas connections academy help me out with physics?

zimovet [89]

I'm not from that school but I can help you.

3 0

3 years ago

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi

USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}\\$ .

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\$ .

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$ .

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$

Now to determine the extension,

$L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\$

4 0

3 years ago

A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as t

salantis [7]

Answer:

2.07

Explanation:

Since you didn't supply the drawing, here is what I assumed:

A is the corner opposite the axis of rotation

B is one of the remaining two corners

L1 is the side between A & B

Centripetal acceleration is given by:

ac = v^2 / r = (v / r) * (v / r) * r…………1

Also angular speed is

w = v / r,………….2

Substituting (2) in (1) gives:

ac = (v / r) * (v / r) * r……….3

= (v / r)^2 * r

= w^2 * r

Therefore, the angular acceleration at A and at B are given by:

acA = w^2 * rA……..4

acB = w^2 * rB……..5

It is given that:

acA = n * acB…………6

Substituting (4) and (5) into (6) gives:

w^2 * rA = n * w^2 * rB ……….7==>

rA = n * rB……..8

In terms of the sides L1 and L2:

rA = sqrt (L1^2 + L2^2)…….9

and

rB = L2…………10

Considering (8):

n * L2 = sqrt (L1^2 + L2^2)………11

Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

Dividing by L2^2:

n^2 = L1^2 / L2^2 + L2^2 / L2^2…….13

= (L1 / L2)^2 + 1 ==>

n^2 - 1 = (L1 / L2)^2 ………14==>

L1 / L2 = sqrt (n^2 - 1) ………15

= sqrt (2.30^2 - 1)

= 2.07. . . . . . <<<=== the value of the ratio L1 / L2 when n = 2.30

8 0

3 years ago