^^ help :c

FrozenT [24]

Hi there!

Great question!

Basketballs have air inside them. A special pump is used to insert the air. That's why you can lift the basketballs off the ground easily. If it was a solid, though, you'd hardly be able to lift the ball up! Basketballs can float, too, because anything with air inside can float. If it were solid, it would sink in the water easily.

Hope this helps! :D

4 0

3 years ago

An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv

konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0

3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

Evgesh-ka [11]

727.5256266 AWNSERrRrrRr

3 0

3 years ago

Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

8 0

3 years ago

Part B

Licemer1 [7]

The number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

<h3>Average daily power consumption</h3>

The average daily power consumption is the amount of electric energy consumed by an individual on a daily rate.

The average daily power consumption of individuals in USA is 12,100 W-hr.

<h3>Converting watts to kilowatts</h3>

E = 12,100 Whr/1000

E = 12.1 kWh

Thus, the number of kilowatts used by an individual to operate his appliances is determined as 12.1 kWh.

Learn more about power here: brainly.com/question/13881533

#SPJ1

5 0

2 years ago