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3 years ago

Need help with this physics question!

Physics

1 answer:

Fynjy0 [20]3 years ago

6 0

Answer:

omg i need help with the same answer lol

Explanation:

i wish i can help but i need help on this hehe

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Which organelle is most like a factory delivery driver?

Fantom [35]

The answer is B, or endoplasmic reticulum.

7 0

3 years ago

Read 2 more answers

10 points

SOVA2 [1]

Explanation:

u=0 m/s

v=9.6 m/s

s= 40.5 m

a=?

V^2=u^2 +2as

9.6^2=0+2×a×40.5

92.16=81a

a=92.16/81

a=1.14m/s^2

3 0

3 years ago

A. What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound f

nasty-shy [4]

Answer:

A)828.8Hz

B)869.2Hz

Explanation:

Here is a complete question;

What frequency is received by a person watching an oncoming ambulance moving at 115 km/h and emitting a steady 753 Hz sound from its siren? Speed of sound is 345m/s

b. What frequency does she receive after the ambulance has passed?

Vs= speed of the ambulance

, We convert to m/s for unit consistency

= 115 km/h= 115km× 1000m/1m × 1hr/3600s= 31.94m/s

Dopler effect is when observed frequency of wave changes with respect to the source or when observed moves relative to transmitting medium can be expressed as

f'=[ (v + vo)/(v- vs)]*f

=[ (v )/(v- vs)]*f

The sign vo and vs depends on vthe direction of the velocity

f= frequency of ambulance siren= 753Hz

v= speed of sound in air= 345m/s

Vo= speed of observer= 0

A) we are to determine the f' of ambulance as heard by person as ambulance approaching.

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v- vs)]*f

=[ (345)/(345-31.94)]×753

= 828.8Hz

B)What frequency does she receive after the ambulance has passed?

To find the frequency f' observed by the person we use the expresion below

Then substitute the values

f'=[ (v )/(v + vs)]*f

=[ (345)/(345 + 31.94)]×753

= 869.2Hz

4 0

2 years ago

If one-third of this energy goes into heat and other forms of internal energy of the motor, with the rest going to the motor out

Naily [24]

Answer

Assuming

electric motor consumes = 9.00 k J

electrical energy for time = 1.00 min

speed of engine = 2500 rpm

Power = $\tau_2\omega_2$

$E = \dfrac{2}{3}\times (P)$

$E = \dfrac{2}{3}\times (9)$

E = 6 k J

E = P t

$P = \dfrac{E}{t}$

$P = \dfrac{6000 J}{60\ s}$

P = 100 W

$\omega_2 = 2600 rpm = 2600 \times \dfrac{2\pi}{60}$

$\omega_2 = 272\ rad/s$

$\tau_2 = \dfrac{P}{\omega_2}$

$\tau_2 = \dfrac{100}{272}$

τ₂ = 0.368 N.m

4 0

3 years ago

The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m2. What is the maximum va

liq [111]

Answer:

The maximum value of the magnetic field in the wave is $7.79\times10^{-8}\ T.$

Explanation:

Given that,

Average value of pointing vector = 0.724 W/m²

We need to calculate the magnetic field

Using formula of pointing vector

$S=\dfrac{1}{\mu_{0}}(E\times B)$

The average value of pointing vector

$S=\dfrac{1}{2\mu_{0}}\times E_{0}B_{0}....(I)$

We know that,

$E_{0}=cB_{0}$

Put the value of $E_{0}$ in equation (II)

$S=\dfrac{1}{2\mu_{0}}\times c(B_{0})^2$

substitute the value in the formula

$0.724=\dfrac{1}{2\times4\pi\times10^{-7}}\times3\times10^{8}\times (B_{0})^2B_{0}=\sqrt{\dfrac{0.724\times2\times4\pi\times10^{-7}}{3\times10^{8}}}B_{0}=7.79\times10^{-8}\ T$

The maximum value of the magnetic field in the wave is $7.79\times10^{-8}\ T.$

4 0

3 years ago