If a data point is way off the trend line which of the following will not help resolve the problem

URGENT

Gemiola [76]

Answer:

5.25 J

Explanation:

W = PE = (f)(x)

PE = 35N*0.15m

PE = 5.25 N*m

1 N*m = 1 J

PE = 5.25 J

7 0

2 years ago

Two cylinders A and B at the same temperature contain the same quantity of the same kind of gas. Cylinder A has three times the

astraxan [27]

Answer:

so pressure in A must be one third the pressure in B

Explanation:

We shall apply gas law to the cylinders A and B . Since their quantity are same so their no of mole will also be same .

For cylinder A

Temperature T , volume 3V , pressure P₁ , no of mole = n

so

P₁ X 3V = n R T

For cylinder B

Temperature T , volume V , pressure P₂ ,no of mole = n

so

P₂ X V = n R T

From the two equation above

P₁ X 3V = P₂ X V

$\frac{P_1}{P_2}=\frac{1}{3}$

P₁ = P₂ / 3

so pressure in A must be one third the pressure in B

3 0

3 years ago

Is Particle B charged or uncharged? If charged, indicate whether it is charged positively or negatively. (n = neutron, p = proto

diamong [38]

Answer:

community College program and maddie d

Explanation:

work business work and maddie

7 0

3 years ago

A coil is wrapped with 300 turns of wire on the perimeter of a circular frame (radius = 8.0 cm). Each turn has the same area, eq

MAVERICK [17]

Answer:

Approximately 18 volts when the magnetic field strength increases from $\rm 20\; mT$ to $\rm 80\;mT$ at a constant rate.

Explanation:

By the Faraday's Law of Induction, the EMF $\epsilon$ that a changing magnetic flux induces in a coil is:

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt}$ ,

where

$N$ is the number of turns in the coil, and
$\displaystyle \frac{d\phi}{dt}$ is the rate of change in magnetic flux through this coil.

However, for a coil the magnetic flux $\phi$ is equal to

$\phi = B \cdot A\cdot \cos{\theta}$ ,

where

$B$ is the magnetic field strength at the coil, and
$A\cdot \cos{\theta}$ is the area of the coil perpendicular to the magnetic field.

For this coil, the magnetic field is perpendicular to coil, so $\theta = 0$ and $A\cdot \cos{\theta} = A$ . The area of this circular coil is equal to $\pi\cdot r^{2} = \pi\times 8.0\times 10^{-2}\approx \rm 0.0201062\; m^{2}$ .

$A\cdot \cos{\theta} = A$ doesn't change, so the rate of change in the magnetic flux $\phi$ through the coil depends only on the rate of change in the magnetic field strength $B$ . The size of the magnetic field at the instant that $B = \rm 50\; mT$ will not matter as long as the rate of change in $B$ is constant.

$\displaystyle \begin{aligned} \frac{d\phi}{dt} &= \frac{\Delta B}{\Delta t}\times A \\&= \rm \frac{80\times 10^{-3}\; T- 20\times 10^{-3}\; T}{20\times 10^{-3}\; s}\times 0.0201062\;m^{2}\\&= \rm 0.0603186\; T\cdot m^{2}\cdot s^{-1}\end{aligned}$ .

As a result,

$\displaystyle \epsilon = N \cdot \frac{d\phi}{dt} = \rm 300 \times 0.0603186\; T\cdot m^{2}\cdot s^{-1} \approx 18\; V$ .

9 0

3 years ago

A 69-kg student sitting on a hardwood floor does not slide until pulled by a 260-N horizontal force. find coefficient of frictio

salantis [7]

Answer:

μ = 0.385

Explanation:

Given that,

The mass of the student, m = 69 Kg

The horizontal force applied, F = 260 N

The normal force acting on the body, weight = mg = 69g N

= 676.2 N

The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.

The formula for coefficient of friction,

μ = F / N

Substituting the values in the above equation,

μ = 260 N / 676.2 N

= 0.385

Hence, the coefficient of friction, μ = 0.385

6 0

3 years ago