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lorasvet [3.4K]
3 years ago
12

Two objects are dropped from a bridge, an interval of 1.00 s apart. What is their separation 1.00 s after the second object is r

eleased
Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

Answer:

<em>171.5m</em>

Explanation:

The velocity of sound in water = 343m/s

Time taken = 1.00secs

using the formula to calculate the distance

2x = vt

x is the distance

v is the speed of sound

t is the time

x = vt/2

x = 343(1)/2

x = 171.5m

<em>hence their separation 1.00 s after the second object is released is 171.5m</em>

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What happens to light waves from a star as the star moves away from Earth?
AURORKA [14]
<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

The astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining the spectra of distant galaxies he observed the spectral lines were displaced towards the red (red shift), whereas the nearby galaxies showed a spectrum displaced to the blue.  

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum. <u>The same happens with the stars and this phenomenom is known as the Doppler effect. </u>

This phenomenon refers to the change in a wave perceived frequency (or wavelength=color) when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.  For example, as a star moves away from the Earth, its espectrum turns towards the red.

8 0
3 years ago
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is
goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0
3 years ago
At which location would you expect the LOWEST TEMPERATURE?
MissTica

Answer:

c

Explanation:

3 0
2 years ago
Read 2 more answers
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
Tatiana [17]

Answer:

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Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

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