Tiles are going to all be lined next to each other, how many inches long will they cover along the walkway?

How do u answer this?

gayaneshka [121]

Answer:

food

Explanation:

did you get a chance to look at the maximum number of devices allowed by

4 0

3 years ago

Why do atoms have an overall neutral charge?

Levart [38]

Https://www.google.com/search?q=Why+do+atoms+have+an+overall+neutral+charge%3F&oq=Why+do+atoms+have+...

Neutrons are no exception. So, if an atom has equal numbers of electrons and protons, the charges cancel each other out and the atom has a neutral charge.

4 0

3 years ago

To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 107 J of energy. Suppose we tried to store th

tatiyna

Answer:

9000RPM

Explanation:

"Angular velocity" is directly related to kinetic energy, that is, the Kinetic energy equation would allow an approximation to the resolution investigated in the problem.

The equation for KE is given by:

$KE = \frac{1}{2} lw ^ 2$

Now, starting from there towards the Angular equation of kinetic energy, the moment of inertia (i) is used instead of mass (m), and angular velocity (w) instead of linear velocity (V)

That's how we get

$KE_{Angular} = \frac{1}{2} Iw^2$

calculating the inertia for a solid cylindrical disk, of

m = 400kg

r = 1.2 / 2 = 0.6m

$I_{disk} = \frac{1}{2} mr^2 = (0.5) (400) (0.6)^2 = 72 kgm^2$

We understand that the total kinetic energy is 3.2 * 10 ^ 7J, like this:

$3.2*10^7 = \frac{1}{2} Iw^2 = (0.5) (72) w^2 = 36w^2$ $w^2 = 3.2*10^7 / 36 = 0.0888*10^7 = 88.8*10^4$

$w = 9.43*10^2 = 943 rad / s$

Thus,

943 rad / s ≈ 9000 rpm

6 0

3 years ago

Please help on this one

Sphinxa [80]

Since this is a projectile motion problem, break down each of the five kinematic quantities into x and y components. To find the range, we need to identify the x component of the displacement of the ball.

Let's break them down into components.

X Y

v₁ 32 cos50 m/s 32 sin50 m/s

v₂ 32 cos50 m/s ?

Δd ? 0

Δt ? ?

a 0 -9.8 m/s²

Let's use the following equation of uniform motion for the Y components to solve for time, which we can then use for the X components to find the range.

Δdy = v₁yΔt + 0.5ay(Δt)²

0 = v₁yΔt + 0.5ay(Δt)²

0 = Δt(v₁ + 0.5ayΔt), Δt ≠ 0

0 = v₁ + 0.5ayΔt

0 = 32sin50m/s + 0.5(-9.8m/s²)Δt

0 = 24.513 m/s - 4.9m/s²Δt

-24.513m/s = -4.9m/s²Δt

-24.513m/s ÷ 4.9m/s² = Δt

5.00s = Δt

Now lets put our known values into the same kinematic equation, but this time for the x components to solve for range.

Δdₓ = v₁ₓΔt + 0.5(a)(Δt)²

Δdₓ = 32cos50m/s(5.00s) + 0.5(0)(5.00)²

Δdₓ = 32cos50m/s(5.00s)

Δdₓ = 102.846

Therefore, the answer is A, 102.9m. According to significant digit rules, neither would be correct, but 103m is the closest to 102.9m so I guess that is what it is.

6 0

3 years ago

g An object with mass m=2 kg is completely submerged, and tethered, to the bottom of a large body of water. If the density of th

Anit [1.1K]

Answer:

The tension in the rope is 20 N

Solution:

As per the question:

Mass of the object, M = 2 kg

Density of water, $\rho_{w} = 1000\ kg/m^{3}$