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Svetllana [295]
3 years ago
9

Electrons are BLANK not Blank -lonic Bond

Physics
1 answer:
DerKrebs [107]3 years ago
4 0

Answer:

electrons are NEGATIVE not POSITIVE ionic bond

Explanation:

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A ball is thrown vertically upward with an initial velocity
ivolga24 [154]

Answer:

D

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

5 0
2 years ago
220V is applied to two different conductors made of the same material. One conductor is twice as long and twice as thick as the
ollegr [7]

Answer:

Explanation:

For calculating resistance of a conductor , the formula is

R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.

For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

R₁ / R₂ = I₁A₂ / I₂A₁

= 2l₂ x A₁ / 4 I₂A₁

= 1 / 2

2R₁ = R₂

Power = V² / R

Ratio of power = (V² / R₁) x (R₂ / V²)

= R₂ / R₁

= 2 .

7 0
3 years ago
In which of these situations is convection most likely the main form of heat transfer? A.Warm air from a heater on the first flo
Sergeeva-Olga [200]
All points ate true

But option A is perfect
6 0
3 years ago
Read 2 more answers
How can potential energy be increased and decreased?
shtirl [24]
By increasing or decreasing the height, we can do that....
4 0
3 years ago
Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show
Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

  1. The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
  2. As the car falls to the ground, its PE converts to KE.
  3. When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

<h3>PE of the car over time</h3>

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height h of the car at time t could be found with the equation:

\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height}),

where

  • g \approx \rm 9.81\; m \cdot s^{-2} near the surface of the earth, and
  • h_0 is the initial height of the car.

On the other hand, \displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}.

In other words, plotting the gravitational PE of the car against time would give a parabola. Since \displaystyle -\frac{m \cdot g^2}{2} < 0 (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at t = 0 the initial GPE is positive, the y-intercept of this parabola should also be positive.

<h3>KE of the car over time</h3>

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's m \cdot g \cdot h_0. If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}.

That's also a parabola when plotted against t. Note that since the quadratic coefficient \displaystyle \frac{m \cdot g^2}{2} is positive, the parabola shall open upwards.

7 0
3 years ago
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