All categories

3 years ago

Riddle of the day What gets wet while drying?

Physics

2 answers:

Ulleksa [173]3 years ago

4 0

Answer:

A towel!

Explanation:

The towel gets wet when it dries something off!

garri49 [273]3 years ago

4 0

Answer:

lol a towel?

Explanation:

because like- I don't know but you get where I'm coming from right?

You might be interested in

Your friend, who is age 15 and has a resting heart rate of 70, wishes to start a jogging program. What would you suggest as a pr

Alex787 [66]

I think the right answer for this question is : 138 because the boy is just 15 year old but he has resting heart rate of 70

6 0

4 years ago

Read 2 more answers

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

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

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

Pls help A ball rolls horizontally off a 100-meter-tall cliff at 40 meters per second. How far does the ball travel horizontally

svet-max [94.6K]

First find the time it takes for the ball to reach the ground using the vertical component of its position vector:

$y=y_0+v_{0y}t+\dfrac12a_yt^2$

$\implies0=100\,\mathrm m+\dfrac12\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

$\implies t=4.52\,\mathrm s$

Meanwhile, the horizontal component of the ball's position vector is

$x=x_0+v_{0x}t+\dfrac12a_xt^2$

$\implies x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)t$

After about 4.52 s, the ball has traveled a horizontal distance of

$x=\left(40\,\dfrac{\mathrm m}{\mathrm s}\right)(4.52\,\mathrm s)=180.8\,\mathrm m$

which you would round to 200 m, so the answer is B.

8 0

3 years ago

Aaron rode her bike to his house to play cards he Road for 1.6 hours at 6.5 KMH another 15 minutes at four KMH what was her aver

Kay [80]

Answer:

The average speed during the trip is 6.16 $\frac{km}{hour}$

Explanation:

Speed is a physical quantity that expresses the variation in position of an object and as a function of time. In other words, speed expresses the relationship between the space traveled by an object, the time used for it and its direction.

The speed can be calculated by the expression:

$Speed=\frac{distance}{time}$

Aaron rode his bike home for 1.6 hours at 6.5 km/h and another 15 minutes at 4 km/h. So, the distance in each of the stages can be calculated as, taking into account that 60 minutes= 1 hour:

distance1=speed*time= 6.5 $\frac{km}{h}$ *1.6 hours= 10.4 km

distance2= speed*time= 4 $\frac{km}{h}$ *15 minutes= 4 $\frac{km}{h}$ *0.25 hours= 1 km

So:

total distance= 10.4 km + 1 km= 11.4 km
total time= 1.6 hours + 0.25 hours= 1.85 hours

Then:

$Speed=\frac{11.4 km}{1.85 hours}$

$Speed=6.16\frac{km}{hour}$

The average speed during the trip is 6.16 $\frac{km}{hour}$

8 0

3 years ago

Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field

ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C

Explanation:

a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

r² = x² + y²

E1 = k q / (x² + y²)

Particle 2. located at x = -24.9 m

r² = x² + y²

E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

cos θ = CA / H = x / (x2 + y2) ½

E1x = E1 cos θ

E2x = = E1 cos θ

The resulting field

Ey = 0

Ex = E1x + E2x 2 E1x

Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

Ex = 497.15 x / (x² + 0.062) ³/₂

Point where can the value of the electric field x = 38.1 cm = 0.381 m

Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂

Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

Ex= 189.41 /0.0943

Ex = 2008 N / C

c) E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

Ex = 497.15 x / (x² + 0.062) ³/₂

Let's use some arithmetic

Ex / 497.15 = x / (x² + 0.062) ³/₂

[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)

The roots of this equation are the solution to the problem,

For Ex = 1.00 kN / C = 1000 N / C

[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

1.312 = (∛x² ) / (x² + 0.062)

1.312 (x² + 0.062) = ∛x²

1.312 X² - ∛x² + 1.312 0.062 = 0

1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0

3 years ago