All categories

3 years ago

What is the difference between storm spotter and storm chaser?

Physics

1 answer:

sveta [45]3 years ago

8 0

Answer:

this is just a guess

Explanation:

a storm spotter only spots the storm and a storm chaser looks for the storm????

You might be interested in

Using the equation for for Newton's Second law, m=F/a solve the following problem. You have been given an object with a force of

inessss [21]

Answer:

4. 5g

Explanation:

F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.

8 0

3 years ago

If the masses are not changed, but the distance of separation is tripled, what happens to the for

Eva8 [605]

If the separation distance between any two objects is tripled (increased by a factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3 raised to the second power).

hope this helps, have a great day
mark brainliest :)!!

7 0

2 years ago

An element's atomic number is 77. How many protons would an atom of this element have? protons

bulgar [2K]

Your answer is 55 protons.

6 0

3 years ago

Read 2 more answers

A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.05 m/s. If the roof is pitched at

vekshin1

1) Time in the air: 0.78 s

The motion of the ball is a projectile motion, which consists of two independent motions:

- A horizontal motion with constant horizontal velocity

- A vertical motion with constant downward acceleration of

$g=-9.8 m/s^2$ (acceleration of gravity)

The initial vertical velocity is

$u_y = u sin \theta = (3.05)(sin(-40^{\circ}))=-1.96 m/s$

where the negative sign means the direction is downward.

The vertical position of the ball is given by

$y(t) = h + u_y t + \frac{1}{2}gt^2$

where

h = 4.50 m is the initial heigth of the ball when it starts falling down

The ball reaches the ground when y = 0, so we have:

$0 = 4.50 -1.96t-4.9t^2$

This is a second-order equation; solving for t, we get

t = -1.18 s

t = 0.78 s

We discard the negative solution since it has no physical meaning, so we can say that the ball spent 0.78 s in the air.

2) Horizontal distance: 1.83 m

For this second part of the problem, we just have to consider the horizontal motion of the ball.

As we said previously, the motion of the ball along the horizontal direction is a uniform motion with constant velocity, which is given by

$v_x = u cos \theta = (3.05)(cos (-40.0^{\circ}))=2.34 m/s$

where u = 3.05 m/s is the initial speed and $\theta$ the angle of projection.

For a uniform motion, we can use the following relationship between distance covered and velocity:

$d=v_x t$

and substituting t = 0.78 s, we find the total distance travelled along the horizontal direction by the ball before reaching the ground:

$d=(2.34)(0.78)=1.83 m$

7 0

3 years ago

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

#SPJ4

5 0

2 years ago