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oksano4ka [1.4K]
3 years ago
12

In an inertia balance, a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the ac

tion of a set of springs. If a 1.00-kg body vibrates at 1.00 Hz, a 2.00-kg body will vibrate at Group of answer choices
Physics
1 answer:
sveticcg [70]3 years ago
4 0

Answer;

a 2.00-kg body will vibrate at 0.707Hz

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How do you think scientists figure out what they think the population will be in 2050?
laila [671]
They would most likely use statistics on the increase/decrease in population from previous decades and centuries. In addition to these statistics, they could use the birth and death rate and ultimately predict the increase within these next 30 years. They must also take climate change and other environmental factors into consideration when formulating such a bold prediction.
4 0
3 years ago
A bus is moving and has 500000 joules of kinetic energy. The brakes are applied and the bus stops. How much work is needed to st
Bad White [126]
For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:
W=K_f - K_i
where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus

Since the bus comes at rest, its final kinetic energy is zero: K_f = 0, so the work done by the brakes to stop the bus is
W=-K_i = -500000 J
And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
8 0
2 years ago
The Hubble Space Telescope has an aperture of 2.4 m and focuses visible light (400-700 nm). The Arecibo radio telescope in Puert
stealth61 [152]

Answer:

y_{hubble} = 77\ \ m

y_{aceribo} = 1.1*10^6 \ \ m

Explanation:

what is the smallest crater that each of these telescopes could resolve on our moon?

For moon ;

s = 3.8 × 10 ⁸ m

y = 1.22 λs/D

where;

λ = 400 nm = 400× 10 ⁻⁹

D = 2.4 m

The smallest crater for the hubble space is calculated as follows:

y_{hubble} = 1.22*400*10^{-9}*3.8*10^8/2.4

y_{hubble} = 77\ \ m

For Aceribo ;

y = 1.22 λs/D

where :

λ = 75 cm = 0.75 m

D = 305 m

y_{acerbo} = 1.22*0.75 *3.8*10^8/305

y_{aceribo} = 1.1*10^6 \ \ m

5 0
3 years ago
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee
lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is -27.3\hat{i}

y component of vector is 43.6\hat{j}

so position vector is

r=-27.3\hat{i}+43.6\hat{j}

Magnitude of vector is

|r|=\sqrt{27.3^2+43.6^2}

|r|=\sqrt{2646.25}

|r|=51.44 units

Direction

tan\theta =\frac{43.6}{-27.3}=-1.597

vector is in 2nd quadrant thus

180-\theta =57.94

\theta =122.06^{\circ}

4 0
3 years ago
A change in which of the following effects the weight of an object?
Masja [62]
Acceleration due to gravity 
4 0
3 years ago
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