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elena55 [62]
3 years ago
8

The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of plane

t 1.
What is the free-fall acceleration on planet 2? Express your answer using two significant figures. m/s^2.
Physics
1 answer:
valkas [14]3 years ago
6 0

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2

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Name:
Brums [2.3K]

Answer:

1.a) 1 kJ

1.b) 4 kJ

     ratio 1:4

1.c) 4 times as before

2.a)  3.33 m/s2

Explanation:

1.a) bicycle's velocity =Displacement/time

                                   =100/20 m/s

                                   =5 m/s

bicycler's KE =1/2 *mass*(velocity)^2

                      =1/2*80*5^2

                       =1000 J = 1 kJ

1.b) bicycle's new velocity =200/20 m/s

                                   =10 m/s

bicycler's new KE =1/2*80*10^2

                             =4000 J = 4 kJ

Ratio= KE 1 :KE new

        = 1 :4

1.c)  when bicycler's speed was doubled it increased the KE by 4 times (2^2). because In KE we consider the square of the speed , so the factor we increase the speed , the KE will get increased with the square value of it

ex : speed is triple the prior value , then the KE is as 3^2 times as before. that is 9 times

2.a) car acceleration = (20-0)/6 m/s2

                                  = 3.33 m/s2

4 0
3 years ago
In a series RLC resonance circuit, the resonance frequency f0 = 700 kHz. The resistor R = 10 Ohm. The specified bandwidth (BW) s
sladkih [1.3K]

Answer:

  • quality factor (Q) = 69.99
  • inductor = 1.591 x 10⁻⁴ H
  • capacitor = 3.248 x 10⁻¹⁰ F

Explanation:

Given;

resonance frequency (F₀) = 700 kHz

resistor, R =  10 Ohm

bandwidth (BW) = 10 kHz

bandwidth (BW)  = \frac{R}{2\pi L}

BW = \frac{R}{2\pi L}

make L (inductor) the subject of the formula

L = \frac{R}{2\pi *BW}  =  \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH

F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}

make C (capacitor)  the subject of the formula

C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F

quality factor (Q) = \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99

quality factor (Q) =  69.99

5 0
3 years ago
A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit
Ratling [72]

Answer:

The magnitude of angular acceleration is 232.38\ rad/s^2.

Explanation:

Given that,

Initial angular velocity, \omega_i=3500\ rev/min=366.5\ rad/s

When it switched off, it comes o rest, \omega_f=0

Number of revolution, \theta=46=289.02\ rad

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2  

So, the magnitude of angular acceleration is 232.38\ rad/s^2. Hence, this is the required solution.

6 0
3 years ago
d is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
Anna71 [15]
<span>D is at rest at the top of a 2 m high slope. The sled has a mass of 45 kg. The sled's potential energy is J?
</span>Answer: The sled's potential energy is 882 Joules
5 0
3 years ago
Read 2 more answers
The presently accepted value of the hubble constant gives an age of
koban [17]

Answer:

14 billion years

Explanation:

The Hubble – Lemaître law, previously called the Hubble law, is a law of physics that states that the redshift of a galaxy is proportional to the distance it is, which is the same as, the further one galaxy is found from another, more quickly it seems to move away from it.

The Hubble constant is the value that measures the rate at which the expansion speed of the Universe varies with distance, and is one of the fundamental parameters of the Universe and allows, in particular, to determine the age of the Universe as we will see.

6 0
2 years ago
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