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3 years ago

Which of the following is the best example of Newton's second law of motion?

Physics

2 answers:

Aloiza [94]3 years ago

7 0

Answer:

Baseball bla

Explanation:

yes

Alex_Xolod [135]3 years ago

3 0

Answer:

C. A baseball bat strikes a baseball.

Explanation:

Newton's second law of motion states that acceleration happens when force is applied on the mass.

<em>A baseball bat strikes a baseball is the best example of Newton's second law of motion because acceleration of the baseball is directly proportional to the force by which baseball bat hit it.</em>

Hence, the correct option is 'C. A baseball bat strikes a baseball.'

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A crate filled with delicious junk food rests on a horizontal floor. Only gravity and the support force of the floor act on it,

stealth61 [152]

The words "... as shown ..." tell us that there's a picture that goes along
with this question, and you decided not to share it. That's sad and
disappointing, but I think the question can be answered without seeing
the picture.

The net force on the crate is zero. Evidence for this is that fact that
the crate is just sitting there. If the net force on an object is not zero,
then the object is accelerating ... it's either speeding up, slowing down,
or its the direction of its motion is changing. If none of these things is
happening, then the net force on the object must be zero.

4 0

3 years ago

9. (02.04 LC)

ycow [4]

Answer:

Molecules speed up

Explanation:

This is caused because of the temperature increasing. The temperature increase is telling us that the thermal energy of the reaction is increasing. When the energy is increased molecules increase their speed, because they have more energy in them

7 0

3 years ago

Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

3 years ago

Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?

lubasha [3.4K]

1) The equivalent resistance of two resistors in parallel is given by:
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}$
so in our problem we have
$\frac{1}{R_{eq}} = \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}$
and the equivalent resistance is
$R_{eq} = \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega$

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
$I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A$

4 0

3 years ago

Read 2 more answers

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

Read 2 more answers