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3 years ago

6

Olivia places her pet frog on a line to observe the frog’s motion. The line is divided into sections that measure 1 centimeter e

ach. The frog begins at 0, moves 18 centimeters forward, moves 6 centimeters backward, and then 12 centimeters backward. What is the frog’s displacement? 0 6 18 36

2 answers:

aalyn [17]3 years ago

4 0

Answer:

Explanation:

Let the forward displacement is taken is positive, and the backward displacement is taken is negative.

first displacement = + 18 cm

second displacement = - 6 cm

third displacement = - 12 cm

net displacement = 18 - 12 - 6 = 0 cm

stepladder [879]3 years ago

4 0

Answer:

A=0

Explanation:

I got 100% on this assignment

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Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magni

Ber [7]

Answer:

a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m

Explanation:

This exercise looks at the motion of a positively charged particle in an electric field.

a) Since the field is vertical the acceleration in this direction is

F = m a

the electric force is

F = q E

we substitute

q E = m a

a = qE / m

the mass of the particle is m = 2.00 10-16 kg

a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg

a = 1,616 m / s²

on the x-axis there are no relationships because there are no forces.

Since the particle has velocities in both axes, its motion is PARABOLIC,

b) the positive particle is accelerated towards the negative plate,

The field is descending, for which the event is down

c) where hit the particle on the x-axis

they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.

Let's find the components of the initial velocity.

sin θ = v_{oy} / v

cos θ = v₀ₓ / v

v_{oy} = v₀ sin θ

v₀ₓ = v₀ cos θ

v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s

v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s

Let's find the time it takes to hit the negative plate

y = y₀I + v_{oy} t + ½ a and t2

in this case the positions are y = y₀ = 0 and the accelerations

a = - 1,616m/s2,

we substitute

0 = 0 + v_{oy} t - ½ a_y t²

v_{oy}= ½ a_y t

t = 2v_{oy} / a_y

let's calculate

t = 2 0.6139 10⁵ / 1.616

t = 7.597 10⁴s

in this time the particle travels a horizontal distance

x = v₀ₓ t

x = 0.8145 10⁵ 7.597 10⁴

x = 6.19 10⁹ m

the particle falls off the plate

4 0

2 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0

oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box $m=2\ kg$

speed of box $v=1.9\ m/s$

distance moved by the box $x=10\ cm$

coefficient of kinetic friction $\mu _k=0.66$

Friction force $f_r=\mu_kN$

$f_r=0.66\times mg$

$f_r=0.66\times 2\times 9.8=12.936 \N$

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

$\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2$

$\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2$

$3.61-1.2936=0.005\times k$

$k=463.28\ N/m$

3 0

2 years ago

Read 2 more answers

Describe the shape of a convex lens and explain what it does to light.

alexira [117]

Convex lenses are thicker at the centers than the edges, they are known as the converging lenses. Rays of light that pass through the lens are brought closer together (they converge). When rays of light that are parallel pass through a convex lens they are refracted, the refracted rays converge at one point called the principal focus.

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3 years ago