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madam [21]
3 years ago
9

HELPPPP PLSSSS A group of students is investigating whether air resistance depends on the size of an object. The students throw

two paper pieces each with the same force. One piece of paper is flat and the other is scrunched into a ball. The distance traveled by each paper piece is shown. Experimental Observations Student 1 Student 2 Student 3 Student 4 Distance traveled by flat paper 0.2 m 1.2 m 1.5 m 0.8 m Distance traveled by scrunched paper 1.5 m 2.4 m 3 m 2 m What was the test variable (independent variable) in this experiment? A.Distance traveled by flat paper B.The shape of the paper C.Distance traveled by scrunched paper D.The number of students
Physics
2 answers:
shutvik [7]3 years ago
8 0

Answer:

Explanation: societal law is A statement about how things act in the natural world. the wire and electro magnet need to touch in question 2. a decrease in spee decreases kinetic energy. the test variable was the distance traveled by the scrunched paper.

juin [17]3 years ago
7 0

Answer:

B

Explanation:

The shape is the only thing that changes.

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To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of
ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= \frac{\mu_0}{4\pi} \frac{2i}{r}

i is current , r is distance from wire

B =  10⁻⁷ x \frac{2\times49}{r}

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

=  10⁻⁷ x \frac{2\times49}{r} x 33 x 1

= 3234 x \frac{10^{-7}}{r}

This should balance weight of second wire per unit length

3234 x \frac{10^{-7}}{r} = .075

r = \frac{3234}{.075} x 10⁻⁷

= .0043 m

= .43 cm .

5 0
3 years ago
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