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3 years ago

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Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a

4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?

2 answers:

Brilliant_brown [7]3 years ago

7 0

Answer:

4.0 kilogramas

Explanation:

gayaneshka [121]3 years ago

6 0

I use the impulse momentum formula.
the 4.0 kilogram ball requires more force to stop

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A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the s

leonid [27]

Complete question is;

A force stretches a wire by 0.60 mm. A second wire of the same material has the same cross section and twice the length.

a) How far will it be stretched by the same force?

b) A third wire of the same material has the same length and twice the diameter as the first. How far will it be stretched by the same force?

Answer:

0.15 mm

Explanation:

According to Hooke's Law,

E = Stress(σ)/Strain(ε)

Where E is youngs modulus

Formula for stress is;

Stress(σ) = Force(F)/Area(A)

Formula for strain is;

Strain(ε) = Change in length/original length = (Lf - Li)/Li

We are also told that a second wire of the same material has the same cross section and twice the length.

Thus;

Rearranging Hooke's Law to get the constants on one side, we have;

F/(AE) = ε

Thus from the conditions given;

ε1 = 0.6/Li

ε2 = (Change in length)/(2*Li)

And ε1 = ε2

Thus;

0.6/Li = Change in length/(2*Li)

Li will cancel out and we now have;

Change in length = 2 × 0.6 = 1.2 mm

Finally, we are told A third wire of the same material has the same length and twice the diameter as the first.

Area of a circle;A1 = πd²/4

Now, we are told d is doubled.

Thus, new area of the new circle is;

A2 = π(2d)²/4 = πd²

Rearranging Hooke's Law,we have;

F/A = εE

Since F and E are now constants, we have;

F/E = constant = Aε

Thus;

A1(ε1) = A2(ε2)

A1 = πd²/4

e1 = 0.60/Li

A2 = πd²

e2 = Change in length/Li

Thus;

((πd²/4) × 0.6)/Li = (πd² × Change in length)/ Li

Rearranging, Li and πd² will cancel out to give;

0.6/4 = Change in length

Change in length = 0.15 mm

4 0

2 years ago

A ball is projected into the air with 100 j of kinetic energy which is transformed to gravitational potential energy at the top

raketka [301]

<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span> $E=K = 100 J$
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>

3 0

3 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

0.0133A

Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

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3 years ago

A 3500 N is force is applied to a spring that has a spring of constant of k= 14000 N/m. How far from equilibrium will the spring

Taya2010 [7]

Answer:

the spring be displaced by 25.0 cm

Explanation:

The computation is shown below:

As we know that

F= -K × x

So,

$x = \frac{-F}{K}$

Now

$x = \frac{-3500}{14000} \\\\$

= -0.250m

= 25.0 cm

Hence, the spring be displaced by 25.0 cm

4 0

3 years ago

Evaporating water produces gaseous water. What has occurred during this process?

Katyanochek1 [597]

Answer:

Heat has accelerated water atoms enough to break the surface tension which leads the liquid to turn into a gas

Explanation:

The state of a substance depends on the distribution of its atoms, therefore any atmosphere change (in this case heat) enough to change the atoms Distribution results in a change of state.

brainliest please ;)

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3 years ago