Are ultrasound waves high energy waves or low energy waves?

Which of the following is not characteristic of a thunderstorm?

blsea [12.9K]

Answer:

option is option d)

Explanation:

the correct option is option d)

the characteristic of thunderstorm can be seen by the following scenario

in thunderstorm there is lot of movement of air the lighter i.e.warmer air moves up and the colder air moves down creating the condition of thunderstorm.

heavy rain , strong wind , potentially large hail stones is the situation of the thunderstorm.

during thunderstorm there is rain and snow at intermediate.

hence only option left is option d.

7 0

3 years ago

Help m-e-e people -_-!

xxTIMURxx [149]

Answer:

the answer is

Explanation:For equilibrium

Weight = Tension

mg=T

∴T=4×3.1π=12.4πN (as can be inferred from the question)

Y=

△l/l

T/A

=

1000

0.031

/20

12.4π/π(

1000

2

)

2

=

4×0.031

12.4×20×1000×(1000)

2

=2×10

12

N/m

2

6 0

3 years ago

Calculate the number of molecules of hydrogen and carbon present in 4 g of methane

jekas [21]

Answer:

Answer = 6.022×10²³ molecules of CH4 which consists of 1 mole of C atoms and 4 moles of H atoms.

3 0

3 years ago

Read 2 more answers

Two particles of equal masses (m = 5.5x10-15 kg) are released from rest with a distance between them is equal to 1 m. If particl

Tcecarenko [31]

To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.

By Coulomb's law, electrical energy is defined as

$EE = \frac{kq_1q_2}{d}$

Where,

EE = Electrostatic potential energy

q= charge

d = distance between the charged particles

k = Coulomb's law constant

While kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m= mass

v = velocity

There by conservation of energy we have that

EE= KE

There is not Initial kinetic energy, then

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2$

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2$

$v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}$

$v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}$

Replacing with our values we have,

$v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}$

$v_f = 2.802*10^7m/s$

Therefore the speed of particle B at the instat when the particles are 3m apart is $2.802*10^7m/s$

4 0

4 years ago

A cannonball is launched horizontally off a 20 m high castle wall with a speed of 85 m/s.

Aneli [31]

Answer:

The ball will be in flight for 2.0 s before it strikes the ground.

The range of the cannonball is 170 m.

Explanation:

Hi there!

The position vector of the cannonball at a time t is given by the following equation:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector of the cannonball at time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Let's place the origin of the frame of reference on the ground, at the edge of the wall so that the initial position vector is r0 = (0, 20) m.

Using the equation of the vertical component of the position vector r, we can find the time it takes the ball to reach the ground:

y = y0 + 1/2 · g · t²

When the cannonball reaches the ground, y = 0:

0 = 20 m - 1/2 · 9.8 m/s² · t²

-20 m / -4.9 m/s² = t²

t = 2.0 s

The ball will be in flight for 2.0 s before it strikes the ground.

Now, we can calculate the horizontal component of the position vector when the ball reaches the ground at t = 2.0 s (i.e. the range of the cannonball).

x = x0 + v0 · t (x0 = 0 because we placed the origin of our frame of reference at the wall).

x = v0 · t

x = 85 m/s · 2.0 s

x = 170 m

The range of the cannonball is 170 m.

3 0

3 years ago