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Sedbober [7]
3 years ago
15

Near the end of a marathon race, the first two runners are separated by a distance of

Physics
1 answer:
yuradex [85]3 years ago
6 0

Answer:

a) 4.75-3.25 = 1.50 m/s

b) Time taken to overtake = 65/1.5 = 43.33 seconds

first runner will finish the race in 215/3.25 = 66.15 seconds

So second runner will win the race

ANSWER

c) 65 +215 = 280m

280 / 4.75 = 58.95 seconds

first runner will complete 58.95 X 3.25 = 191.5875 m

So 215-191.5875=23.4125 meters ahead

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A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving
Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0
3 years ago
Use this formula to solve this problem: Power (P) = Work (W) ÷ time (t) Peter's body supplies a force of 500 N to run up a 5 met
sesenic [268]
As you said p=w/t.
but, w=f×s
=500×5=2500j
t=10s
p=w/t
=2500/10=250 watts
4 0
3 years ago
A pulley system is used at a dock to lift shipments of fish off a boat. If you apply a force of 100 N to the pulley, it pulls th
labwork [276]

Explanation:

a. Mechanical advantage = force out / force in

MA = 830 / 100

MA = 8.3

b. Efficiency = work out / work in

0.80 = W / 600 J

W = 480 J

7 0
3 years ago
An experiment is performed aboard the International Space Station to verify that linear momentum is conserved during collisions
Anni [7]

Answer:

v=15.9554\ m.s^{-1}

Explanation:

Given:

  • mass of the honey drop 1, m_1=35.5\times 10^{-3}\ kg
  • velocity of the honey drop 1, v_1=13.1\ m.s^{-1}
  • mass of the honey drop 2, m_1=52.3\times 10^{-3}\ kg
  • velocity of the honey drop 2, v_2=14.5\ m.s^{-1}
  • mass of the honey drop 3, m_1=75.7\times 10^{-3}\ kg
  • velocity of the honey drop 3, v_3=18.3\ m.s^{-1}

<em>In ISS there is zero gravity an the collision is completely inelastic.</em>

<u>So, applying the law of conservation of momentum:</u>

m_1.v_1+m_2.v_2+m_3.v_3=(m_1+m_2+m_3).v

35.5\times 10^{-3}\times 13.1+52.3\times 10^{-3}\times 14.5+75.7\times 10^{-3}\times 18.3=(35.5+52.3+75.7)\times 10^{-3}\times v

v=15.9554\ m.s^{-1}

4 0
3 years ago
At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's
FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s          

Explanation:

Use first equation of motion to find components of velocity at a given time:

v = u + at

where, v is  the final velocity, u is the initial velocity, a is the acceleration and t is the time.

Given:

u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s

v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s

v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s

5 0
3 years ago
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