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3 years ago

Please answer with an explanation.

Physics

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

A, D, E

Explanation:

The force of Earth's gravitational pull is equal to the mass of the object times the acceleration due to gravity:

W = mg

The greater the mass, the greater the pull. A truck, bicycle, and camel all have greater mass than a baseball.

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which of the following is not a property of an acid? reacts with metals bitter taste reacts with carbonate turns litmus paper re

notka56 [123]

The property that does not describe an acid is that it has a bitter taste. Acids have a sour taste, Bases have a bitter taste.

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4 years ago

Which is the best aerobic exercise plan?

Arisa [49]

Answer:

Running or Jogging

Running and jogging are both great options for aerobic conditioning. Whether you run at the gym or outside, you are in control of setting the intensity of your workout. When aiming to build muscle mass, you can add more resistance or jog at an incline, along with increasing your speed.

Explanation:

6 0

4 years ago

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An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe —

sattari [20]

Answer:

r = 2,026 10⁹ m and T = 2.027 10⁴ s

Explanation:

For this exercise let's use Newton's second law

F = m a

where the force is electric

F = $k \frac{q_1q_2}{r^2}$

Acceleration is centripetal

a = v² / r

we substitute

$k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}$

r = $k \frac{q_1q_2}{m \ v^2}$ (1)

let's look for the charge in the insulating sphere

ρ = q₂ / V

q₂ = ρ V

the volume of the sphere is

v = 4/3 π r³

we substitute

q₂ = ρ $\frac{4}{3}$ π r³

q₂ = 3 10⁻⁹ $\frac{4}{3}$ π 4³

q₂ = 8.04 10⁻⁷ C

let's calculate the radius with equation 1

r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)

r = 2,026 10⁹ m

this is the radius of the electron orbit around the charged sphere.

Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio

v = x / t

the distance traveled in a circle is

x = 2π r

In this case, time is the period

v = 2π r /T

T = 2π r /v

let's calculate

T = 2π 2,026 10⁹/628 103

T = 2.027 10⁴ s

4 0

3 years ago

A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a tube of length

lora16 [44]

Answer:

T = 4.42 10⁴ N

Explanation:

this is a problem of standing waves, let's start with the open tube, to calculate the wavelength

λ = 4L / n n = 1, 3, 5, ... (2n-1)

How the third resonance is excited

m = 3

L = 192 cm = 1.92 m

λ = 4 1.92 / 3

λ = 2.56 m

As in the resonant processes, the frequency is maintained until you look for the frequency in this tube, with the speed ratio

v = λ f

f = v / λ

f = 343 / 2.56

f = 133.98 Hz

Now he works with the rope, which oscillates in its second mode m = 2 and has a length of L = 37 cm = 0.37 m

The expression for standing waves on a string is

λ = 2L / n

λ = 2 0.37 / 2

λ = 0.37 m

The speed of the wave is

v = λ f

As we have some resonance processes between the string and the tube the frequency is the same

v = 0.37 133.98

v = 49.57 m / s

Let's use the relationship of the speed of the wave with the properties of the string

v = √ T /μ

T = v² μ

T = 49.57² 18

T = 4.42 10⁴ N

4 0

3 years ago

Read 2 more answers

A uniform rod with a mass of 100 g and a length of 50.0 cm rotates in a horizontal plane about a fixed vertical, frictionless pi

Alex787 [66]

(a) The angular speed of the system at the instant the beads reach the end of the rod is 9.26 rad/s.

(b) The angular speed of the rod after the after the beads fly off the rod's ends is 25.71 rad/s.

<h3>Moment of inertia through the center of the rod</h3>

I = ¹/₁₂ML²

I = ¹/₁₂ (0.1)(0.5)²

I = 0.0021 kgm²

For the beads, I = 2Mr² = 2(0.03 x 0.1²) = 0.0006 kgm²

Total initial moment of inertia, Ii = 0.0021 kgm² + 0.0006 kgm²

I(i)= 0.0027 kgm²

When the beads reach the end, I = 2Mr² = 2(0.03)(0.25)² = 0.00373 kgm²

Total final moment of inertia, I(f) = 0.0021 kgm² + 0.00373 kgm²

I(f) = 0.00583 kgm²

<h3>Speed of the system</h3>

The speed of the system at the moment the beads reach the end of the rod is calculated as follows;

$I_i \omega_i = I_f\omega _f\\\\\omega _f = \frac{I_i \omega_i}{I_f}$

$\omega_f = \frac{0.0027 \times 20}{0.00583} \\\\\omega _f = 9.26 \ rad/s$

<h3>Speed of the rod when the beads fly off</h3>

$\omega_f = \frac{0.0027 \times 20}{0.0021} \\\\\omega _f = 25.71 \ rad/s$

Learn more about moment of inertia of rods here: brainly.com/question/3406242

5 0

3 years ago