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2 years ago

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A 16-kg scooter is moving at a speed of 7 m/s. The scooter’s speed doubles. What is the scooter’s kinetic energy when its speed

doubles?
a. 784 J
b. 1568 J
c. 392 J
d. 112 J

1 answer:

lbvjy [14]2 years ago

5 0

The formula for kinetic energy is:

KE = 0.5mv^2

where:
m = mass
v = speed

Given this formula, the original KE of the scooter is calculated as 392 J. Since the speed is doubled, it can be observed in the formula that the change would actually affect the kinetic energy by quadrupling it. This is because 2^2 is 4. So, 392(4) = 1568 J or B.

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A cliff diver running 3.60 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2.00 s later

mart [117]

Explanation:

It is given that,

The horizontal speed of a cliff diver, $v_x=3.6\ m/s$

It reaches the water below 2.00 s later, t = 2 s

Let $d_x$ is the distance where the diver hit the water. It can be calculated as follows :

$d_x=v_x\times t\\\\=3.6\times 2\\\\=7.2\ m$

Let $d_y$ is the height of the cliff. It can be calculated using second equation of motion as follows :

$d_y=u_yt+\dfrac{1}{2}gt^2\\\\d_y=\dfrac{1}{2}\times 9.8\times 2^2\\\\=19.6\ m$

So, the cliff is 19.6 m high and it will hit the water at a distance of 19.6 m.

8 0

1 year ago

for a given initial projectile speed, you observe that the projectile has a certain range R at a launch angle of a = 30. For wha

VLD [36.1K]

Answer:

The other angle is 30 degrees.

Explanation:

The range of projectile is given by :

$R=\dfrac{u^2\ \sin2\theta}{g}$

Here,

u is the speed of launch of projectile

Here, $\theta=30^{\circ}$

We need to find the other launch angle when the projectile have the same range, such that,

$\dfrac{u^2\ \sin(60)}{g}=\dfrac{u^2\ \sin2\alpha}{g}$

$\sin(60)=\sin2\alpha$

$\dfrac{\sqrt3}{2}=\sin2\alpha$

$\alpha =30^{\circ}$

So, the other angle is 30 degrees. Hence, this is the required solution.

3 0

2 years ago

A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the studen

Brums [2.3K]

We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
$\frac{h_i}{h_0}=\frac{d_i}{d_0}$
Two triangles that I marked are similar and from this we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f}$
The image and the object must have the same height so we get:
$\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\
1=\frac{d_i-f}{f}\\
d_i=2f$
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
$S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm$

8 0

1 year ago

The model of the atom has changed as scientists have gathered new evidence. Four models of the atom are shown below, but one imp

nexus9112 [7]

Answer: Dalton’s model

Explanation:

In the attached image we can see four atomic models labeled with four letters:

W represents the current and accepeted atomic model: a nucleus with an electron cloud, where the orbit and position of the electrons around the nucleus is defined by specific regions (associated with specific energy levels) where there is a greater probability of finding the electron at any given moment. It is important to note this model was improved by the works in quantum physics done by Louis de Broglie and Erwin Schrodinger.

X represents Rutherford's model (This model was proposed after Thomson's model). Ernest Rutherford conducted a series of experiments in order to corroborate Thomson's atomic model. However the results of the experiment led him to find out there is a concentration of charge in the atom's core (which was later called nucleus) surrounded by electrons. This lead to a new atomic model, in which the atom has a positive charged nucleus surrounded by negative charged particles that move similar to the orbit of the planet around the Sun.

Y represents Thomson's model, also called the <em>plum pudding</em> model. This scientific found out that atoms contain small subatomic particles with a negative charge (later called electrons). However, taking into consideration that at that time there was still no evidence of the atom nucleus, Thomson thought the electrons were immersed in the atom of positive charge that counteracted the negative charge of the electrons. Just like the raisins embedded in a pudding or bread.

Z represents Bohr's model. This model was proposed by the danish physicist Niels Bohr after Rutherford's model. In fact, this model was Rutherford's model with the following addition: electrons orbit the nucleus (like planets around the sun) in specific orbits at different energy levels around the nucleus.

So, the only missing model is <u>Dalton's model</u>, which was the first atomic model: the atom represented as a solid, indestructible and indivisible mass. An idea that was already accepted by that time since the ancient Greeks.

4 0

2 years ago

Read 2 more answers

Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12

Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

$-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-$

In case all three capacitors are connected in parallel, the $2.0\;\mu\text{F}$ capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

$-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-$

In case the $2.0\;\mu\text{F}$ capacitor is connected in parallel with the $1.5\;\mu\text{F}$ capacitor, and the two capacitors in parallel is connected to the $3.0\;\mu\text{F}$ capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}$ .

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

$Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}$ .

Voltage is the same across two capacitors in parallel.As a result,

$\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}$ .

<h3>Case C: 2.76 V.</h3>

$-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-$ .

Similarly,

the effective capacitance of the two capacitors in parallel is 5.0 μF;
the effective capacitance of the three capacitors, combined: $\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}$ .

Charge stored:

$Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}$ .

Voltage:

$\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}$ .

<h3 /><h3>Case D: 4.00 V</h3>

$-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-$ .

Connect all three capacitors in series.

$\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}$ .

For each of the three capacitors:

$Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}$ .

For the $2.0\;\mu\text{F}$ capacitor:

$\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}$ .

6 0

1 year ago