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2 years ago

5) Choose the correct statement about the waves shown below.

Physics

1 answer:

abruzzese [7]2 years ago

4 0

Answer:

the answer is b sorry if its wrong

Explanation:

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If two objects of the same size move through the air at different speeds, which encounters the greater air resistance?a. The fas

blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

Speed of the object
Cross-sectional area of the object
Shape of the object

Formula:

$F=\frac{1}{2}C_d\rho A v^{2}$

As the speed of the object increases the amount of Air resistance/drag increases on the object, as the above formula shows direct relation between Air resistance/drag and velocity i.e F ∝ v^2.

6 0

3 years ago

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.1-m-long rope. The ball is pulled to one s

AlladinOne [14]

Answer:

The tension in the rope is 262.88 N

Explanation:

Given:

Weight $W = 150$ N

Length of rope $r = 4.1$ m

Initial speed of ball $v = 5.5 \frac{m}{s}$

For finding the tension in the rope,

First find the mass of rod,

$mg = 150$ ( $g = 9.8 \frac{m}{s^{2} }$ )

$m = \frac{150}{9.8}$

$m = 15.3$ kg

Tension in the rope is,

$T = mg + \frac{mv^{2} }{r}$

$T = 150 + \frac{15.3 \times (5.5)^{2} }{4.1}$

$T = 262.88$ N

Therefore, the tension in the rope is 262.88 N

7 0

3 years ago

A solar cooker, really a concave mirror pointed at the sun, focuses the sun's rays 16.0 cm in front of the mirror. what is the r

VARVARA [1.3K]

The focal point of a concave mirror is halfway along the radius, therefore the radius would be 2•16= 32 cm

5 0

2 years ago

Which statement is not true about radiation

Mekhanik [1.2K]

I don’t know i’m sorry

4 0

2 years ago

In the diagram, q1 = +6.39*10^_9 C and

ivanzaharov [21]

Answer:

The electric potential will be "259.695 volt".

Explanation:

In the given question, the figure is not provided. Below is the attached figure given.

Given:

$q_1=6.39\times 10^{-9} \ C$

$q_2=3.22\times 10^{-9} \ C$

$AP=(0.150+0.250)$

$=0.40 \ m$

$BP=0.25 \ m$

Now,

At point P, the electric potential will be:

⇒ $V=\frac{q_1}{4 \pi \epsilon_o AP } +\frac{q_2}{4 \pi \epsilon_o BP}$

By putting values, we get

⇒ $=9\times 10^9 [\frac{6.39\times 10^{-9}}{0.40} +\frac{3.22\times 10^{-9}}{0.25} ]$

⇒ $=259.695 \ Volt$

4 0

3 years ago