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Elis
3 months ago
13

Jeff puts on a leather jacket over his sweater. The sweater becomes negatively charged. Which statements about Jeff’s situation

are true?
The sweater has a tendency to attract electrons.

The sweater has a tendency to attract protons.

The leather jacket has a tendency to attract electrons.

The leather jacket has a tendency to attract protons.

The leather jacket has a lower tendency to attract electrons than the sweater.


NEED ASAP!!! GOTTA FINISH BEFORE NEXT CLASS
Physics
2 answers:
inna [3.1K]3 months ago
8 0

Response:

A and E

Explanation:

Sav [3.1K]3 months ago
7 0

The following statements hold true
(i) The leather jacket is less likely to attract electrons than the sweater.
When the sweater and the leather jacket come into contact, the leather jacket loses electrons and becomes positively charged, while the sweater gains electrons and thus becomes negatively charged.
Opposite charges attract, meaning the sweater (which is negatively charged) will pull in protons (which are positively charged). The leather jacket (positively charged) will in turn attract electrons (which are negatively charged).

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It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you
Maru [3345]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

F_X = F_{cos \ \theta}

Substituting 42° for θ and 87.0° for F

F_X =87.0 \ N \ *cos \ 42 ^\circ

F_X =64.65 \ N

Meanwhile, the vertical component is:

F_Y = Fsin \ \theta

Again substituting 42° for θ and 87.0° for F

F_Y =87.0 \ N \ *sin \ 42 ^\circ

F_Y =58.21 \ N

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

f = \mu \ N

Where;

\mu is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting F_Y \ for \ F_{sin \ \theta}. Normal reaction becomes:

N = mg \ - \ F_Y

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

F_X = \mu \ ( mg - \ F_Y)

Rearranging the equation above to isolate \mu leads to:

\mu \ = \ \frac{F_X}{mg - F_Y}

Substituting in the following values:

F_X \ = \ 64.65 \ N

m = 73 kg

g = 9.8 m/s²

F_Y = \ 58.21 N

Thus:

\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}

\mu = 0.0984

Therefore, the coefficient of friction is = 0.0984

5 0
3 months ago
If the axes of the two cylinders are parallel, but displaced from each other by a distance d, determine the resulting electric f
serg [3582]

Response:

E =  ρ ( R1²) / 2 ∈o R

Clarification:

Provided information

Two cylinders are aligned parallel

Distance = d

Radial distance = R

d < (R2−R1)

To determine

Express the response using the variables ρE, R1, R2, R3, d, R, and constants

Solution

We have two parallel cylinders

therefore, area equals 2 \pi R × l

And we apply Gauss's Law

EA = Q(enclosed) / ∈o......1

Initially, we calculate Q(enclosed) = ρ Volume

Q(enclosed) = ρ ( \pi R1² × l )

Thus, inserting all values into equation 1

produces

EA = Q(enclosed) / ∈o

E(2 \pi R × l)  = ρ ( \pi R1² × l ) / ∈o

This simplifies to

E =  ρ ( R1²) / 2 ∈o R

6 0
2 months ago
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