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3 years ago

5

Many hard-of-hearing people like sound compressed because they remain sensitive to ________ sounds. prolonged high-pitched frequ

ent unpredictable loud

1 answer:

pychu [463]3 years ago

7 0

Answer:

loud

Explanation:

Noise exposure results in a decrease in the sensitivity of auditory sensory cells thereby affecting the hearing ability of the concerned.

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An object is placed in front of a thin converging lens of focal length 12 cm. The image of the object is upright and 2.5 times a

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3 years ago

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3 years ago

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II).

aleksley [76]

Answer:

- There was a constant acceleration at 0 to 10s

- There was a zero acceleration at 10 to 25s

- There was a constant deceleration at 25 to 30s

Explanation:

<em>See attachment for complete question.</em>

Solving (a): What happens at 0s to 10s

There was a constant acceleration and this is proven below.

At time 0, velocity = 15

At time 10, velocity = 30

This is represented as:

$(t_1,v_1) = (0,15)$

$(t_2,v_2) = (10,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-15}{10 - 0}$

$A = \frac{15}{10}$

$A = 1.5$

<em>Since the acceleration is positive, then it shows a constant acceleration.</em>

Solving (b): What happens at 10s to 25s

There was a zero acceleration and this is because the velocity do not change.

See proof below

At time 10, velocity = 30

At time 25, velocity = 30

This is represented as:

$(t_1,v_1) = (10,30)$

$(t_2,v_2) = (25,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-30}{25 - 10}$

$A = \frac{0}{15}$

$A = 0$

Solving (c): What happens at 25s to 30s

There was a constant deceleration and this is proven below.

At time 25, velocity = 30

At time 30, velocity = 0

This is represented as:

$(t_1,v_1) = (25,30)$

$(t_2,v_2) = (30,0)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{0-30}{30-25}$

$A = \frac{-30}{5}$

$A = -6$

<em>Since the acceleration is negative, then it shows a constant deceleration</em>

4 0

3 years ago

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

lianna [129]

Answer: E. 450 K

Explanation:

It is given that the temperature of the gaseous planet is linearly related with height of the atmosphere. we can write this in the mathematical equation:

y = m x +c

where y is the temperature values, x is height, m is the slope and c is the y-intercept. we have been given two sets of value in the image, using which we can find the value of slope in y-intercept.

at x = 18.40 km, y = 147.54 K

⇒147.54 = 18.40 m + c

⇒c = 147.54 - 18.40 m ..(1)

at x =78.11 km, y = 567.00 K

⇒567.00 =78.11 m + c ..(2)

Put equation 1 in 2 and solve:

⇒567.00 =78.11 m + 147.54 - 18.40 m

⇒419.46 = 59.71 m

⇒ m =419.46 ÷59.71 = 7.025 K/km

c = 147.54 - 18.40 × 7.025 = 18.28 K

At height, x = 61.5 km the approximate temperature is :

y = 7.025 K/km × 61.5 km + 18.28 K = 450.3 K

Thus, the approximate temperature at altitude 61.5 km is 450 K.

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4 years ago