Why does it take much longer time and distance to stop a moving ship in water than a moving car on the road

Complete the passage to describe the relationship between kinetic energy, internal energy, thermal energy, and

butalik [34]

Answer:

Increases

Explanation:

I don't know if you answered your own question but I'll just answer this for others confused ahh

4 0

2 years ago

Read 2 more answers

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

or

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

or

$cos X=0.8951$

or

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

or

$cos Y=0.649$

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

2 years ago

This is a velocity versus time graph of a car starting from rest. If the area under the line is 10 meters, what is the correspon

adelina 88 [10]

So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.

So this appears to be a right triangle so we can find the area of a triangle as:

0.5bh = A

Since our area is 10 meters lets alter our formula a bit to fit the situation:

Our base here is time and our height is velocity so:

0.5tv = Δx

So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

So we have v, and Δx so lets isolate for time by dividing by v and 0.5

t = Δx / 0.5v

Now lets plug all that in:

t = 10 / 0.5(2)

t = 10 seconds

Hope this helped!

8 0

3 years ago

A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r

Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0

2 years ago

90cm uniform lever has a load 30N suspended at 15cm from one of it's end. If the fulcrum is at the center of gravity. The force

solniwko [45]

Answer: F = 20 N

Explanation:

I will ASSUME that the fulcrum is at the center of gravity of the lever arm, This means that the lever arm itself creates no moment about the fulcrum because there is no moment arm for that particular force.

To solve, we sum moments about any convenient point to zero (zero because there is no acceleration in the F = ma equation)

The easiest convenient point is the fulcrum

30((90/2) - 15) - F(90/2) = 0

30(30) = F(45)

F = 900/45 = 20 N

3 0

2 years ago