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4 years ago

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suppose s(t) is the position of an object moving along a line at time t is greater than or equal to 0. What is the average veloc

ity between the times t=a and t=b? What is the instantaneous velocity at t=a?

1 answer:

allsm [11]4 years ago

8 0

<span>I think the average velocity b/w t =a and t= b would be
Avg vel. = change in vel./ change in time</span>

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Un carro tiene una velocidad angular de 95 rev/min. a) ¿Cuál es la velocidad tangencial del objeto si se colocan unos contrapeso

Maru [420]

Answer:

v = 5.949 m / s , a = 59.18 m / s²

Explanation:

Linear and angular quantities are related

v = w r

we search the magnitudes to the SI system

w = 95 rev / min (2π rad / 1rev) (1 min / 60 s) = 9.948 rad / s

d = 598 mm = 0.598 m

a) let's calculate the linear velocity

v = 9.948 0.598

v = 5.949 m / s

b) Cenripetal acceleration

a = v² / r

a = 5.94 2 / 0.598

a = 59.18 m / s²

6 0

4 years ago

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an elect

klio [65]

Complete Question

An electron is accelerated by a 5.9 kV potential difference. das (sd38882) – Homework #9 – yu – (44120) 3 The charge on an electron is 1.60218 × 10−19 C and its mass is 9.10939 × 10−31 kg. How strong a magnetic field must be experienced by the electron if its path is a circle of radius 5.4 cm?

Answer:

The magnetic field strength is $B= 0.0048 T$

Explanation:

The work done by the potential difference on the electron is related to the kinetic energy of the electron by this mathematical expression

$\Delta V q = \frac{1}{2}mv^2$

Making v the subject

$v = \sqrt{[\frac{2 \Delta V * q }{m}] }$

Where m is the mass of electron

v is the velocity of electron

q charge on electron

$\Delta V$ is the potential difference

Substituting values

$v = \sqrt{\frac{2 * 5.9 *10^3 * 1.60218*10^{-19} }{9.10939 *10^{-31]} }$ f

$= 4.5556 *10^ {7} m/s$

For the electron to move in a circular path the magnetic force[ $F = B q v$ ] must be equal to the centripetal force[ $\frac{mv^2}{r}$ ] and this is mathematically represented as

$Bqv = \frac{mv^2}{r}$

making B the subject

$B = \frac{mv}{rq}$

r is the radius with a value = 5.4cm = $= \frac{5.4}{100} = 5.4*10^{-2} m$

Substituting values

$B = \frac{9.1039 *10^{-31} * 4.556 *10^7}{5.4*10^-2 * 1.60218*10^{-19}}$

$= 0.0048 T$

7 0

4 years ago

Match the organisms to the descriptions.

makvit [3.9K]

Answer:

Lacks colored blood - Scorpion

Soft, unsegmented body - Octopus

A vertebrate - Bird

Lacks antennae - Starfish

Explanation:

Scorpion belongs to the Phylum Arthropoda of the Kingdom Animalia. All the organisms of this phylum lack coloured blood.

Octopus comes under the Phylum Molusca of the Kingdom Animalia. Soft, unsegmented body is the property of the organisms of this phylum.

Bird belongs to the Vertebrata class of the Phylum Chordata of Kingdom Animalia.

Starfish belongs to Echinodermata Phylum of the Kingdom Animalia. The organisms of this category do not posses antennae.

6 0

4 years ago

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At a time when mining asteroids has become feasible, astronauts have connected a line between their 3740-kg space tug and a 5690

Gwar [14]

Answer:

61.4 s

Explanation:

The distance d₁ traveled by the asteroid:

$d_1=\frac{1}{2}a_1t^2$

The distance d₂ traveled by the space ship:

$d_2=\frac{1}{2}a_2t^2$

The total distance d:

$d=d_1+d_2=\frac{1}{2}(a_1+a_2)t^2$

Solving for time t:

$t=\sqrt{\frac{2d}{a_1+a_2}}=\sqrt{\frac{2d}{F(\frac{1}{m_1}+\frac{1}{m_2})}}=\sqrt{\frac{2dm_1m_2}{F(m_1+m_2)}}$

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3 years ago

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A swimmer does 3560 j of work in 55 s what is the swimmers power output?

ankoles [38]

Explanation:

<em>Here </em><em>it </em><em>is </em><em>given </em>

<em>Work </em><em>(</em><em>W) </em><em> </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em> </em><em>J</em>

<em>Time </em><em>(</em><em>t) </em><em> </em><em>=</em><em> </em><em>5</em><em>5</em><em> </em><em>sec</em>

<em>power </em><em>(</em><em>P) </em><em> </em><em>=</em><em> </em><em>?</em>

<em>We </em><em>know </em><em>we </em><em>have </em><em>the </em><em>formula </em>

<em> $p = \frac{w}{t}$ </em>

<em>P </em><em>=</em><em> </em><em>3</em><em>5</em><em>6</em><em>0</em><em>/</em><em>5</em><em>5</em>

<em>P </em><em>=</em><em> </em><em>6</em><em>4</em><em>.</em><em>7</em><em>3</em><em> </em><em>watt</em>

8 0

3 years ago