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frozen [14]
3 years ago
13

Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at

rest. After he lands on the raft, the raft, with him on it, moves away from the pier at 1.6 m/s. What was Tyrone's speed as he ran off the end of the pier?
Physics
1 answer:
Ghella [55]3 years ago
4 0

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

mu=(M+m)v

85\times u=(85+130)\cdot 1.6

u=\frac{215}{85}\cdot 1.6

u=2.529\cdot 1.6=4.04 m/s  

You might be interested in
A 0.450 kg soccer ball has a kinetic energy of 119 J.
Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2  = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0
3 years ago
5. Calculate the acceleration of a 2 kg block across a table if you push with a force of 20 N and the frictional force is 4 N.
frozen [14]

Answer:

Explanation:hey do you go to GOC too?

8 0
2 years ago
What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still
Inessa [10]

Answer:

physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

3 0
3 years ago
An object weighing 49 N is dropped from a height of 30 m. It is found to be moving with a velocity of 24m/s just before it hits
faltersainse [42]

The  frictional force will be 0.22N.

<h3>What is Frictional force?</h3>

Frictional force is the force generated between two surfaces that are in contact and slide against each other.

Given,

Weight=4N

mass =4.9/9.8=0.5kg

Hieght =30m

velocity=24m/s

Acceration , v²-u²=2as

24²/2×30 =a , u is zero

 a= 1.5m/s²



By Using conservation  of energy ,

30F+1/2mv²=mgh

30F=1150-144

F= 6/30

F=0.2N

The force will be 0.2N

to learn more about Friction clickhttps://brainly.com/question/17608236

#SPJ9

6 0
2 years ago
A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsul
jeka94

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

<h3>Given:</h3>

Mass of the astronaut, m_a = 126 kg

Speed he acquires, v_{a}  = 2.70 m/s

Mass of the space capsule, m_{c} = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

P_f = m_av_a + m_cv_c

P_I = 0

m_av_a + m_cv_c = 0

v_c =\frac{- m_a v_a}{m_c}}\\\\

   = \frac{126* 2.70}{1800}

   = - 0.189 m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2} × 126 × (2.70) ^2

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = \frac{1}{2} m v^2

     = \frac{1}{2}×1800×(0.189) ^2

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

#SPJ1

3 0
2 years ago
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