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2 years ago

A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average

Physics

1 answer:

Kaylis [27]2 years ago

3 0

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

2 years ago

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

$\lambda = 0.186 m$

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

$f' = f_0 (\frac{v + v_o}{v - v_s})$

here we know that

$f_0 = 1750 Hz$

$v_o = 15 m/s$

$v_s = 15 m/s$

now we will have

$f' = (1750)(\frac{340 + 15}{340 - 15})$

$f' = 1911.5 Hz$

Part b)

Apparent wavelength is given by the formula

$\lambda = \frac{v_{relative}}{f_{app}}$

here we will have

$\lambda = \frac{340 + 15}{1911.5}$

$\lambda = 0.186 m$

3 0

2 years ago

In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we

jekas [21]

Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is $w_{h}$ .

We need to calculate the weight of heart of a human

Using given data

$w_{h}=0.5\times w$

Where, h = weight of heart

w = weight of human

$w_{h}=\dfrac{0.5}{100}\times 185$

$w_{h}=0.93\ lbs$

Hence, The weight of heart of a human is 0.93 lbs.

8 0

2 years ago

Dawn and Aram have stretched a slinky between them and begin experimenting with waves. As the frequency of the waves is doubled

m_a_m_a [10]

Answer:

halved

Explanation:

The velocity of the a wave is obtained by multiplying the frequency and wavelength.

$v=f\lambda\\\Rightarrow f=\frac{v}{\lambda}\\\Rightarrow \lambda=\frac{v}{f}$

Where

v = Velocity

f = Frequency

$\lambda$ = Wavelength

The velocity here is constant. So, if the frequency is doubled the wavelength is halved.

6 0

2 years ago

In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal

Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

p = m1 v1 + m2 v2

Part B.

Find the derivative

dp / dt = m1 dv1dt + m2 dv2 / dt

dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1 f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0

2 years ago