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3 years ago

Freeeeeeee points!!!

Physics

2 answers:

TiliK225 [7]3 years ago

8 0

Answer:

Thanks

Explanation:

OMG! My name is MAYA too!!!!!!!!!!!!!YAY!!!

I'm gonna friend request u

Andrej [43]3 years ago

7 0

Answer:

thanks so much lol can I get brainliest

Explanation:

You might be interested in

Which type of relationship is formed when a megabat eats a fig and drops the

Alina [70]

The type of relationship formed when a megabat eats a fig and drops the seeds in a new location is COMENSALISM. It is an ecological interaction.

<h3>What is commensalism?</h3>

Commensalism is a type of ecological interaction in which one organism benefits (in this case, the tree) and the other organism neither benefits nor harm (the megabat).

Mutualism is a type of ecological association in which both organisms benefit from such interaction.

Conversely, parasitism is a type of ecological interaction in which one organism benefits and the other organism is harmed.

Learn more about commensalism here:

brainly.com/question/16712254

3 0

2 years ago

30 points plz help ill do anything... literally anything.

ruslelena [56]

Answer:

1. 2.5s

Explanation:

1. For time, divide Distance / speed

25m / 10

=2.5s

3 0

3 years ago

The brick wall (of thermal conductivity 0.35 W/m ·◦ C) of a building has dimensions of 2.7 m by 6 m and is 16 cm thick. How much

Gwar [14]

Answer:

$Q = 54.577\,MJ$

Explanation:

The heat transfer through brick wall is:

$\dot Q = \frac{k\cdot A}{L}\cdot \Delta T$

$\dot Q = \frac{\left(0.35\,\frac{W}{m\cdot ^{\circ}C} \right)\cdot (2.7\,m)\cdot (6\,m)}{0.16\,m} \cdot (31^{\circ}C - 8^{\circ}C)$

$\dot Q = 815.063\,W$

The heat flow in a 18.6-h period is:

$Q = \dot Q \cdot \Delta t$

$Q = (815.063\,W)\cdot (18.6\,h)\cdot \left(\frac{3600\,s}{1\,h} \right)$

$Q = 54576618.48\,J$

$Q = 54.577\,MJ$

8 0

3 years ago

Read 2 more answers

What is the value of R2 in this parallel circuit? (5 stars)

o-na [289]

Answer:

20 Ω

Explanation:

Voltage, current, and resistance are related by Ohm's law:

V = IR

40 V = (4 A) R

R = 10 Ω

The total resistance of the circuit is 10 Ω.

Resistors in parallel have a total resistance of:

1/R = 1/R₁ + 1/R₂

1 / (10 Ω) = 1 / (20 Ω) + 1/R₂

R₂ = 20 Ω

4 0

2 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago