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3 years ago

Which of the following would NOT be a medium that mechanical waves move through

Physics

1 answer:

MaRussiya [10]3 years ago

3 0

Answer:

i don't have any answer without the options

Explanation:

Where is the pic

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Problem 1: Problem 1.58 in Young & Freedman A plane leaves the airport in Galisteo and flies 170 km at 68.0° east of north;

Maurinko [17]

Answer:

Explanation:

We shall convert the displacement in vector form using unit vector i and j

consider east as x axis and north as y axis

170 km at 68.0° east of north

D₁ = 170 sin68 i + 170cos 68 j

= 157 i + 63.68 j

230 km at 36.0° south of east

D₂ = 230 cos36 i - 230 sin 36 j

= 186 i - 135.2 j

Resultant Displacement = D₁ +D₂

= 157 i + 63.68 j + 186 i - 135.2 j

= 343 i - 71.52 j

Resultant magnitude

= √ ( 343² + 71.52²

= 350 km

Angle

= tan⁻¹ ( - 71.52 / 343 )

12⁰ south of east .

4 0

3 years ago

Someone please help. science <3 thx<br> ill give 15 pts for it.

STALIN [3.7K]

Answer:

put these numbers in the boxes from up to down. hope this helps! :)

Explanation:

5 0

3 years ago

What is the number of daylight hours of daylight difference in the summer for two regions located at different latitudes in the

ruslelena [56]

To be completely honest the answer is 74

4 0

3 years ago

A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri

Alla [95]

Answer:

$q=1\times10^{-8}C$

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

$F=\frac{KQq}{d^{2}}$

By substituting the values

$0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}$

$q=1\times10^{-8}C$

Thus, the charge on the ball bearing is $q=1\times10^{-8}C$

7 0

3 years ago

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Naya [18.7K]

Answer:

$2.966\times 10^{-11}\ N$

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = $6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}$

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

$F=\dfrac{GMm}{d^2}$

Plug in the given values and solve for 'F'. This gives,

$F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})$

The same force is experienced by both cannonball and marble.

Therefore, the gravitational force of the marble is $2.966\times 10^{-11}\ N$

3 0

3 years ago