A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .
Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.
Given
time = 20 μs = 20 * 
s
let distance to the reflecting surface be = x
total distance travelled by pulse will be = 2x
speed = 3.0 × 
m/s
distance = speed * time
2x = 3.0 × * 20 *
x = 3000 m
The reflecting object would be 3000 m away
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Answer:
Im not 100% sure but i think the answer is A. An electron in an atom jumping from a lower energy state to a higher one.
Explanation:
lmk if its wrong
Answer:
Specific heat of brass is 0.40 J g⁻¹ °C⁻¹ .
Explanation:
Given :
Mass of brass, m₁ = 440 g
Temperature of brass, T₁ = 97° C
Mass of water, m₂ = 350 g
Temperature of water, T₂ = 23° C
Specific heat of water, C₂ = 4.18 J g⁻¹ °C⁻¹
Equilibrium temperature, T = 31° C
Let C₁ be the specific heat of brass.
Heat loss by brass = Heat gain by water
m₁ x C₁ x ( T₁ -T ) = m₂ x C₂ x ( T - T₁ )
Substitute the suitable values in above equation.
440 x C₁ x (97 - 31) = 350 x 4.18 x (31 - 23)
C₁ = 
C₁ = 0.40 J g⁻¹ °C⁻¹
Answer:
Explanation:
Due to heat energy , metal expands . Formula for linear expansion is as follows .
L = l ( 1 + α Δt )
where L is expanded length , l is original length , α is coefficient of linear expansion and Δt is increase in temperature .
To pass the sphere through the ring , the diameter of both ring and sphere should be same after heating . Let after increase of temperature Δt , their diameter becomes same as L . The linear coefficient of brass and steel are
20 x 10⁻⁶ and 12 x 10⁻⁶ respectively .
For steel sphere ,
L = 25 ( 1 + 12 x 10⁻⁶ Δt )
For brass ring
L = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
25 ( 1 + 12 x 10⁻⁶ Δt ) = 24.9 ( 1 + 20 x 10⁻⁶ Δt )
1.004( 1 + 12 x 10⁻⁶ Δt ) = ( 1 + 20 x 10⁻⁶ Δt )
1.004 + 12.0482 x 10⁻⁶ Δt = 1 + 20 x 10⁻⁶ Δt
.004 = 7.9518 x 10⁻⁶ Δt
Δt = 4000 / 7.9518
= 503⁰C.
final temp = 503 + 15 = 518⁰C .
