A scientific theory is:

What does the "coefficient of friction" tell you?

Arte-miy333 [17]

a coefficient of fraction is a value that shows the relationship between two objects and the normal reaction between the objects that are involved.

8 0

3 years ago

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into it. The roller coaster starts

avanturin [10]

Answer:

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

Explanation:

In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:

K.E at A + P.E at A = K.E at B + P.E at B

(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)

(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)

where,

vₙ = velocity of roller coaster at point a = 0 m/s

hₙ = height of roller coaster at point a = 25 m

g = 9.8 m/s²

vb = velocity of roller coaster at point B = ?

hb = Height of Point B = 0 m (since, point is the reference point)

Therefore,

(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)

245 m²/s² * 2 = vb²

vb = √(490 m²/s²)

<u>So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.</u>

5 0

3 years ago

A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between

Sunny_sXe [5.5K]

Answer : $4.483\times 10^{15}\ m/s^2$ .

Explanation:

It is given that,

Electric field strength, $E=2.55\times 10^{4}\ N/C$

We know that,

Charge of electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

From the definition of electric field, $F=qE$ ...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

$ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}$

$a=0.4483\times 10^{16}\ m/s^2$

or

$a=4.483\times 10^{15}\ m/s^2$

So, the horizontal component of acceleration of an electron is $4.483\times 10^{15}\ m/s^2$ .

Hence, it is the required solution.

7 0

3 years ago

A car is on a circular off ramp of an interstate and is traveling at exactly 25 mph around the curve. Does the car have velocity

netineya [11]

Answer:

The car has velocity and acceleration but is not decelerating

Explanation:

Since the car is traveling at 25 mph around the curve, it has a tangential velocity. This tangential velocity is constantly changing in direction (so the car could adapt to the curve and not moving forward in a straight line), there should be a centripetal acceleration in play here. This acceleration does not slow down the car so it's not decelerating.

6 0

3 years ago

Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure

Lynna [10]

Answer:

yes it was a constant speed and the car traveled 10 meters in 20 seconds.

Explanation:

8 0

3 years ago