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Over [174]
3 years ago
8

A dryer sheet is attracted to your hand as your pull laundry from the dryer. Which of the following statements is true? Select a

ll that apply.
1. The dryer sheet is positively charged and your hand is negatively charged.
2. The dryer sheet is negatively charged and your hand is positively charged.
3. With the given information it is not possible to tell which charge either object has.
4. The dryer sheet and your hand have opposite charges.
5. The dryer sheet and your hand have the same charge.
Physics
2 answers:
bixtya [17]3 years ago
8 0

Answer:

2. The dryer sheet is negatively charged and your hand is positively charged.

4. The dryer sheet and your hand have opposite charges.

Explanation:

As we know that electrostatic force between two charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that two charges here are of similar type i.e. either both are positive or both are negative then in that case the force between them is repulsive force and if the charges are of opposite sign then the electrostatic force between two charges are attraction type of force.

So here If our hand is attracted towards the dryer sheet then our hand and dryer sheet must have opposite charges due to which they are attracted.

Also we know that the electron affinity of dryer sheet is more due to which it has tendency not to lose the electron so here we can say dryer sheet is negatively charge and hand is positively charged

guapka [62]3 years ago
7 0
2. The dryer sheet is negatively charged and your hand is positively charged
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If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?
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The car's angular speed is \frac{rad}{s}.

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3 0
3 years ago
A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo
horrorfan [7]

Given that,

Mass of object = 9.90 kg

Time =5.40 s

Suppose the force is (2.00i + 9.00j + 5.30k) N, initial position is (2.70i - 2.90j + 5.50k) m and final position is (-4.10i + 3.30j + 5.40k) m.

We need to calculate the displacement

Using formula of displacement

s=r_{2}-r_{1}

Where, r_{1} = initial position

r_{2} = final position

Put the value into the formula

s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)

s= -6.80i+6.20j-0.1k

(a). We need to calculate the work done on the object

Using formula of work done

W=F\cdot s

Put the value into the formula

W=(2.00i + 9.00j + 5.30k)\cdot (-6.80i+6.20j-0.1k)

W=-13.6+55.8-0.53

W=41.67\ J

(b). We need to calculate the average power due to the force during that interval

Using formula of power

P=\dfrac{W}{t}

Where, P = power

W = work

t = time

Put the value into the formula

P=\dfrac{41.67}{5.40}

P=7.71\ Watt

(c). We need to calculate the angle between vectors

Using formula of angle

\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})

Put the value into the formula

\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})

\theta=79.7^{\circ}

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

(c).  The angle between vectors is 79.7°

7 0
3 years ago
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