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2 years ago

7

Identify the traces on the oscilloscope screen and explain what happens to the pitch of the sound in both A and B. Mention your

answers clearly by writing A and B.

1 answer:

jolli1 [7]2 years ago

5 0

Answer:

the assessment is due at 11 you only have 1 hour do it fasst

Explanation:

the assessment is due at 11 you only have 1 hour do it fasst

You might be interested in

How do i calculate this?

Lesechka [4]

CAR 1

Momentum = Mass/Velocity

M = 2100/20

M = 105 m/s^2

CAR 2

Momentum = Mass/Velocity

M = 2100/30

M = 70 m/s^2

8 0

3 years ago

what is the average speed of (a) a car that travels 400m in 20s. and (b) an athlete who runs 1500m in 4 minutes

zmey [24]

Answer:

a) 20 m/s

b) 37.5 m)s

Explanation:

Average speed = total distance ÷ total time

=> (a) average speed of a car that travels 400m in 20s

= 400/20 = 20 m/s

& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)

= 1500/240 = 37.5 m/s

5 0

3 years ago

The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to

DedPeter [7]

Answer:

$9.4\cdot 10^{10} m$

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

$\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}$

where

$r_o$ is the distance of the new object from the sun (orbital radius)

$T_o=180 d$ is the orbital period of the object

$r_e = 1.50\cdot 10^{11} m$ is the orbital radius of the Earth

$T_e=365 d$ is the orbital period the Earth

Solving the equation for $r_o$ , we find

$r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m$

3 0

3 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago

A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is

soldier1979 [14.2K]

W =F triangle d cosine0. F = 25 Newton’s. Delta d = 50 meters. Theta =40.0 degrees

3 0

3 years ago