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Debora [2.8K]
3 years ago
15

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp

eed remained the same, by what factor would its centripetal acceleration change?
Physics
1 answer:
laila [671]3 years ago
7 0

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

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Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha weighs 637 N, and each boot has a mass of 4.50 kg. A
Studentka2010 [4]

Answer:

-0.73 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, in absence of external forces (the ice is frictionless, so no friction), the total momentum of Aisha + two boots is conserved.

At the beginning, their total momentum is zero, since they are at rest:

p_i = 0 (1)

After, their total momentum is:

p_f = Mv + 2mv' (2)

where:

M is Aisha's mass

v is Aisha's velocity relative to the ground

m = 4.50 kg is the mass of each boot

v' is the boot's velocity relative to the ground

We can find:

M=\frac{W}{g}=\frac{637 N}{9.8 N/kg}=65 kg is Aisha's mass (where W = 637 N was her weight)

v' can be rewritten as:

v'=v+6

because 6 m/s is the velocity of the boots relative to her, while v' is their velocity relative to the ground.

Substituting and combining (1) and (2) we find:

0=Mv+2m(v+6)\\0=Mv+2mv+12m\\v=\frac{-12m}{M+2m}=\frac{-12(4.50)}{65+2(4.50)}=-0.73 m/s

and the negative sign indicates that the direction is opposite to that of the boots.

8 0
3 years ago
A. Blue light has higher energy than red light. Write 3 - 4 sentences comparing these electromagnetic waves with respect to the
frosja888 [35]

PART A)

As we know that energy of light depends on its wavelength and frequency as following formula

E = \frac{hc}{\lambda} = h\nu

now we know that wavelength of blue light is less than the red light so here energy of blue light will be more

also we know that

\nu = \frac{c}{\lambda}

so here if wavelength is smaller for blue light so its frequency will be high and the speed of both light will be same in same medium

PART B)

Since we know that frequency of blue light is more than red light as well as wavelength of blue light is less than the wavelength of blue light so here blue light will have more energy

When blue light and red light strike the metal surface then due to more energy of blue light it will release some loosely bonded electrons from metal surface which will contribute in current.

here if we increase the intensity of light then the number of photons that contain the blue light of certain energy will be more and that will contribute more current

So here quantification help as we know that due to quantization only certain frequency or energy will lead to eject electron so all colours will not give this current

5 0
3 years ago
Read 2 more answers
An ultrasound unit is being used to measure a patient's heartbeat by combining the emitted 2.0 MHz signal with the sound waves r
alexandr402 [8]
Hi there, 
for this question we have:
Signal 2.0 MHz = Emitted so we can call it f_e
and we need the Reflected = f_{r}
In this question, we have a source which goes to the heart and a reflected which comes back from the heart and we need the speed of the reflected.
So you should know that the speed of reflected is lower than the source(Emitted). 
we also know: ΔBeat frequency(max) = 560 Hz = f_{b}
so we have: 
f_{e} - f_{r} = f_{b}
so frequency of Reflected is: 
2.0 × 10^6 Hz - 560 Hz = 1.99 × 10^6 Hz = f_{r}
now you know that Lambda = v/f 
so if we find the lambda with our Emitted then we can find v with the Reflected: 
Lambda = 1540(m/s) / 2.0 × 10^6 Hz = 7.7 × 10^-4 m 
=> v_{max} = (lambda)(f_{r} 
=> 7.7 × 10^-4m (1.99 × 10^6Hz) = 1532 m/s 
so the v_{max} is equal to 1532 m/s :)))
This question is solved by two top teachers as fast as they could :))
I hope this is helpful
have a nice day

8 0
3 years ago
1. What is the kinetic energy of a 94 kg basketball player running at 7.3 m/s?
LUCKY_DIMON [66]
1. 2504.63 J

2. 929.5m/s

Hope this helps :)

I can explain it as well if you’d like
8 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
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