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Rainbow [258]
2 years ago
9

Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his

house to talk to a neighbor. It takes Ben 17 minutes from the time he leaves his house until he stops to talk to his neighbor.
What is Ben’s average velocity?

0.2 m/s
0.5 m/s
0.7 m/s
1.2 m/s

Physics
2 answers:
ehidna [41]2 years ago
8 0

Answer:

velocity = 0.2 m/s

Explanation:

As we know that Ben walks to the market at distance of 500 m

then he turns back towards his house and continue 200 m past his house towards neighbor.

So the return path length is 700 m

So here total displacement of Ben is given as

d = 700 - 500

d = 200 m

Now for the average velocity we know that

velocity = \frac{displacement}{time}

velocity = \frac{200}{17 \times 60 seconds}

velocity = 0.2 m/s

bonufazy [111]2 years ago
3 0

(answer with work in image)

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aniked [119]

Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

  • v is the speed of the spaceship
  • d is the distance
  • t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996  light-years

This distance is<em> very slightly less than 4 light-years. </em>

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2 years ago
You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force
Tpy6a [65]

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

F = Force Applied = 120 N

a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

<u>Mass of Little Sister = 44.17 kg</u>

4 0
2 years ago
Astronauts in the International Space Station must work out every day to counteract the effects of weightlessness. Researchers h
Snowcat [4.5K]

Answer:

  33.725 rpm

Explanation:

The relationship between rotational speed in radians per second and acceleration is ...

  \omega=\sqrt{\dfrac{a}{r}}

We want the rotation rate in RPM, so we need the conversion ...

  \text{RPM}=\dfrac{\text{rad}}{\text{s}}\cdot\dfrac{1\,\text{rev}}{2\pi\,\text{rad}}\cdot\dfrac{60\,\text{s}}{1\,\text{min}}

Then the required rotational speed in RPM is ...

  RPM=\sqrt{\dfrac{a}{r}}\cdot\dfrac{30}{\pi}=\dfrac{30}{\pi}\sqrt{\dfrac{1.4\cdot 9.8}{1.1}}\approx 33.725

The rotation rate needs to be about 33.7 rpm to give an acceleration of 1.4g at the astronaut's feet.

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2 years ago
A motorcyclist heading east through a small Iowa town accelerates after he passes a signpost at x=0 marking the city limits. His
adoni [48]

Answer:

1) v_2=23\ m.s^{-1}              &     x_2=43\ m east of sign post

2) x'=55\ m east of sign post

3) x_n=205\ m east of the signpost.

4) v_z=35\ m.s^{-1}

Explanation:

Given:

  • position of motorcyclist on entering the city at the signpost, x_0=0\ m
  • time of observation after being at x=5m east of the signpost, t_m=0\ s
  • constant acceleration of the on entering the city, a=4\ m.s^{-2}
  • distance of the motorcyclist moments later after entering, s_m=5\ m
  • velocity of the motorcyclist moments later after entering, u_m=15\ m.s^{-1}

<u>Now the initial velocity on at the sign board:</u>

u_m^2=u^2+2.a.x_m

where:

u= initial velocity of entering the city at the signpost

Putting respective values:

15^2=u^2+2\times 4\times 5

u=13.6015\ m.s^{-1}

1)

Position at time t_2=2\ s sec.:

Using equation of motion,

x_2=u_m.t_2+\frac{1}{2} a.(t_2)^2+5 because it has already covered 5m before that point

x_2=15\times 2+0.5\times 4\times 2^2+5

x_2=43\ m east of sign post

Velocity at time t_2=2\ s sec.:

v_2=u_m+a.t_2

v_2=15+4\times 2

v_2=23\ m.s^{-1}

2)

Position when the velocity is v'=25\ m.s^{-1}:

using equation of motion,

v'^2=u_m^2+2.a.x'+5

25^2=15^2+2\times 4\times x'+5

x'=55\ m east of sign post

3)

Given that:

acceleration be, a_n=2\ m.s^{-2}

time, t_n=5\ s

Position after the new acceleration and the new given time:

using equation of motion,

x_n=u_m.t_n+\frac{1}{2} a_n.(t_n)^2+5

x_n=15\times 5+0.5\times 2\times 5^2+5

x_n=205\ m east of the signpost.

4)

now time of observation, t_z=5\ s

v_z=u_m+a.t_z

v_z=15+4\times 5

v_z=35\ m.s^{-1}

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lana66690 [7]
I'm thinking that you're supposed to divide. So you would divide 5 into 60 and get 12
6 0
1 year ago
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