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3 years ago

QUICK!! What class of leer is shown below?

Physics

2 answers:

stiv31 [10]3 years ago

8 0

The fourth one. I’m not sure but it’s highlighted in blue so...

wlad13 [49]3 years ago

5 0

Answer:

3rd class lever

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As the frequency of a wave increases the wavelength.

zaharov [31]

Answer:

decreases

Explanation:

Remeber:

There is always inverse relation between frequency and wavelength.

So if one of them increases, other decreases and vice-versa.

f ∝ 1 / λ

4 0

2 years ago

Read 2 more answers

When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Star

Gemiola [76]

Answer:

α = 200*π rad/sec²

Explanation:

If the disk drive revolves once every 0.05 sec, this means that the angular velocity can be expressed as follows:

$\omega = \frac{2*\pi}{0.05s} = 40*\pi rad/sec$

(just applying the definition of angular velocity)
Now, it the disk started from rest, and took two revolutions to reach to full speed, assuming that the angular acceleration is constant, we can find the angular acceleration applying this kinematic equation:

$\omega_{f} ^{2} -\omega_{0} ^{2} = 2*\alpha * \Delta\theta$

where Δθ, is the angle rotated while it went from rest to full speed, i.e., 2 revolutions, or 2*π rads.
Replacing by the value that we found for ωf (as ω₀ = 0), we can solve for α :

$\alpha =\frac{\omega_{f}^{2} }{2*\Delta\theta} = \frac{(40*\pi)^{2} }{2*2*2*\pi } =\\ \\ \alpha = 200*\pi rad/sec2$

3 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

a 8000kg tow truck pulls a 3000kg car horizontally into a garage. if the motor of the tow truck exerts a forward thrust on the t

Evgen [1.6K]

Equation 1 :
m1 : m1a =T
Equation 2:m2 : m2a= F - T

Adding 1 and 2a=0.073

Placing in equation 1

we get T = 218.18N

8 0

3 years ago

In order to place a satellite into orbit, it requires enough fuel to supply the necessary mechanical energy. Into what types of

nexus9112 [7]

When a satellite is revolving into the orbit around a planet then we can say

net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have

$F_g = F_c$

$\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}$

now we can say that kinetic energy of satellite is given as

$KE = \frac{1}{2}M_s v^2$

$KE = \frac{GM_sM_p}{2r}$

also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it

so we will have

$U = -\frac{GM_sM_p}{r}$

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite

6 0

3 years ago