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V125BC [204]
3 years ago
10

6. A lumberjack is standing on a log floating on a lake. She starts from rest, then runs along the log to the end, when she jump

s from the first log onto a second. After landing safely on the second log, she slows down and ends up standing on the second log. Both logs both have masses of 150 kg each and the mass of the lumberjack is 70 kg. The lumberjack reaches a speed of 7.0 m/s relative to the shore during her jump. What are the speeds of the two logs and of the lumberjack after she has stopped on the second log
Physics
1 answer:
scoray [572]3 years ago
4 0

Answer:

a) -3.267 m/s

b) 2.227 m/s

Explanation:

As per the conservation of momentum

m1v1 + m2v2=0

m1= mass of log

m2 = mass of lumber jack

v1 = velocity of log

v2 = velocity of lumber jack

a) Velocity of first log

-\frac{70*7}{150} = -3.267 m/s

b) m1v1 + m2v2 = m3v3

Velocity of log

= \frac{70*7}{150+70} \\2.227

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A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
3 years ago
Read 2 more answers
If the force on a hammer is 24 N and its mass is 1.6 kg, then the
Gnesinka [82]

Answer:

15 m/s²

Explanation:

F = ma

make "a" the subject

a = F/m

6 0
3 years ago
Select ALL of the following statements that provide evidence that there is friction acting on a cart moving along a level track.
Sati [7]

Answer:

1st and 4th one................

6 0
3 years ago
Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

  = 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

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

u = 0

So that,

v = 14 x 510

 = 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

 = 7 x 260100

 = 1820700 m

The distance that the shuttle has traveled during the given time is  1820.7 Km.

5 0
3 years ago
A light wave has a wavelength of 450 nanometers. What is the frequency of this light?
Zolol [24]

Answer:

Frequency, f=6.67\times 10^{14}\ Hz

Explanation:

Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,

c=f\lambda

f is the frequency of this light

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{450\times 10^{-9}}\\\\f=6.67\times 10^{14}\ Hz

So, the frequency of this light is 6.67\times 10^{14}\ Hz.

3 0
3 years ago
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