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2 years ago

Probably the deadliest aspect of a thunderstorm is _____. rain thunder lightning hail

2 answers:

max2010maxim [7]2 years ago

8 0

The answer to the given question is lightning.
Lightning is most probably the deadliest aspect of a thunderstorm. It is a sudden electrostatic discharge during an electrical storm. It happens when there is an electrostatic discharge between electrically charged regions of a cloud, which is referred as the intra-cloud lightning or IC, between that cloud and another cloud or the CC lightning, or between a cloud and the ground (CG lightning).

kipiarov [429]2 years ago

3 0

Probably the deadliest aspect of a thunderstorm is lightning .

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Which are the two most popular candidates for gamma-ray bursters? Group of answer choices collisions between a white dwarf and a

kow [346]

Answer:

hypernova making a black hole, and merger of two neutron stars

7 0

2 years ago

a 72kg person is standing on a bathroom scale (calibrated in newtons). what is the reading of the scale when the acceleration is

Likurg_2 [28]

Answer:

F=ma

therefore A=F/M

Explanation:

i think that's what your doing but I'm not sure

4 0

2 years ago

Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w

PSYCHO15rus [73]

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

8 0

3 years ago

Read 2 more answers

PLEASE HELP ASAP!!!!

sveticcg [70]

Answer:

it's frequency

Explanation:

hope this helps :)

8 0

3 years ago

Read 2 more answers

A 20-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 26 n. starting from rest, the sle

Vesna [10]

Good morning.

We calculate the acceleration with the Torricelli equation:

$\star\ \boxed{\mathsf{V^2 = V_0^2 + 2a\Delta S}}$

We see that:

$\begin{cases}\mathsf{V_0 = 0}\\\mathsf{V = 2.3 \ m/s}\\\mathsf{\Delta S = 8.9 \ m}\end{cases}$

Now:

$\mathsf{(2.3)^2 = 0^2 + 2a\cdot 8.9}\\ \\ \mathsf{5.29 = 17.8a}\\ \\ \bold{\mathsf{a = 0.3 \ m/s^2}}$

Now we can calculate the resultant force that makes that acceleration of 0.3 m/s² with the 2nd Law of Newton:

$\mathsf{F_r = m\cdot a}\\ \\ \mathsf{F_r = 20\cdot 0.3}\\ \\ \mathsf{F_r = 6 \ N}$

We have a force of 26 N → and a friction force F ←. Adding those vectors, he have a force 6 N →. Therefore:

26 - F = 6

F = 20 N

We have a friction force of 20 N. We calculate the kinect coefficient with the formula:

$\star \ \boxed{\mathsf{F = \mu_k N}}$

Since we are in a horizontal plane, we hava that N = P = mg = 200 N

Therefore:

$\mathsf{20 = \mu_k 200}\\ \\ \boxed{\boxed{\mathsf{\mu_k = 0.1}}}$

5 0

3 years ago