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3 years ago

Question 8 of 25

Physics

1 answer:

GuDViN [60]3 years ago

7 0

Answer:

A. They use X-rays that penetrate most objects.

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3. What will happen to the black bass and blue gill as the floor of the ponds fills with organic

yKpoI14uk [10]

Answer: die

Explanation: oyxagan all goon bc of all dat suffs

8 0

3 years ago

How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a spa

Zarrin [17]

Answer:

cutting the magnet in two parts each part has a North and South pole,

Explanation:

In magnetism the magnetic mono-poles are not found, this means that we do not have magnetic charges alone, therefore when cutting the magnet in two parts each part has a North and South pole, the magnetic lines go from the North pole to the South pole, see attached.

The density of the lines is approximately the intensity of the magnetic field.

4 0

3 years ago

Give an example for each of the following, where the force:

max2010maxim [7]

Explanation:

A cricket player hitting the ball from opposite direction.
A footballer kicking ball with more force.
Heating of a plastic bottle.
Applying brake of a car.
rolling a stopped marble on a table.

7 0

2 years ago

Read 2 more answers

Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir le

kap26 [50]

Answer:

Explanation:

1 ha = 10⁴ m²

1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²

In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³

Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m

Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³

Let Q be the withdrawal in m³

Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶

Q = 26.20 x 10⁶ m³

rate of withdrawal per second

= 26.20 x 10⁶ / 30 x 24 x 60 x 60

= 26.20 x 10⁶ / 2.592 x 10⁶

= 10.11 m³ / s

6 0

3 years ago

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

ziro4ka [17]

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

$d = 0.5 \times 9.8 \times {4.2}^{2}$
on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

3 0

3 years ago