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3 years ago

13

The graph depicts the velocity and times of Elan and Anna during a race.

2 answers:

Ratling [72]3 years ago

8 0

Answer:

C

Explanation:

vovangra [49]3 years ago

3 0

Answer: C

Explanation:

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I have my exam monday I need some help!

Lana71 [14]

Answer:

Explanation:

If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.

The distance between the 2 crests (or troughs) is the wavelength, λ.

We want the velocity of this wave. The equation that relates these 3 things is

$f=\frac{v}{\lambda}$ and filling in:

$2.5=\frac{v}{2.0}$ so

v = 2.5(2.0) and

v = 5.0 m/s

7 0

3 years ago

PLS ANSWER FAST WILL MARK BRAINLIEST

densk [106]

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7 0

3 years ago

Two spherical asteroids have the same radius R. Asteroid 1 has mass M and asteroid 2 has mass 1.97·M. The two asteroids are rele

nekit [7.7K]

Answer:

$0.536\sqrt{\frac{GM}{R}}$

Explanation:

We are given that

Mass of one asteroid 1, $m_1=M$

Mass of asteroid 2, $m_2=1.97 M$

Initial distance between their centers,d=13.63 R

Radius of each asteroid=R

d'=R+R=2R

Initial velocity of both asteroids

$u=0$

We have to find the speed of second asteroid just before they collide.

According to law of conservation of momentum

$(m_1+m_2)u=m_1v_1+m_2v_2$

$(M+1.97 M)\times 0=Mv_1+1.97Mv_2$

$Mv_1=-1.97 Mv_2$

$v_1=-1.97v_2$

According to law of conservation of energy

$Gm_1m_2(\frac{1}{d'}-\frac{1}{d})=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2$

$GM(1.97M)(\frac{1}{2R}-\frac{1}{13.63R})=\frac{1}{2}M(-1.97v_2)^2+\frac{1}{2}(1.97M)v^2_2$

$1.97M^2G(\frac{13.63-2}{27.26R})=\frac{1}{2}Mv^2_2(3.8809+1.97)$

$1.97MG(\frac{11.63}{27.26 R})=\frac{1}{2}(5.8509)v^2_2$

$v^2_2=\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}$

$v_2=\sqrt{\frac{1.97GM\times11.63\times 2}{27.26R\times 5.8509}}$

$v_2=0.536\sqrt{\frac{GM}{R}}$

Hence, the speed of second asteroid = $0.536\sqrt{\frac{GM}{R}}$

8 0

3 years ago

Clouds, wind, and rain are part of the _____.

Musya8 [376]

A its Stratosphere, Sorry I didn't see your answer, its bilogy I think not physics.. :)

8 0

3 years ago

A centrifuge in a medical laboratory rotates at an angular speed of 3500 rev/min. When switched off, it rotates 46.0 times befor

BlackZzzverrR [31]

Answer:

The angular acceleration of the centrifuge is -231.74 rad/s².

Explanation:

Given that,

Angular speed = 3500 rev/min = 366 rad/s

We need to calculate the angular displacement

Using formula of angular displacement

$\theta=2\pi n$

Put the value into the formula

$\theta=2\pi\times46$

$\theta=289.02\ rad$

We need to calculate the angular acceleration

Using equation of motion

$\omega_{f}^2=\omega_{i}^2+2\alpha\theta$

$\alpha=\dfrac{\omega_{f}^2-\omega_{i}^2}{2\theta}$

$\alpha=\dfrac{0-(366)^2}{2\times289.02}$

$\alpha=-231.74\ rad/s^2$

Hence, The angular acceleration of the centrifuge is -231.74 rad/s².

3 0

3 years ago