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dlinn [17]
3 years ago
6

How long would a 2950 N force need to act to cause a 6000 kg delivery truck to accelerate from 4.0m/s to 29.0m/s?​

Physics
1 answer:
AysviL [449]3 years ago
8 0

Answer:

Time, t = 50.8 seconds.

Explanation:

Given the following data;

Mass = 6000 kg

Force = 2950 N

Initial velocity = 4 m/s

Final velocity = 29 m/s

To find time;

First of all, we would determine the acceleration of the truck using the following formula;

Force = mass * acceleration

2950 = 6000 * acceleration

Acceleration = 2950/6000

Acceleration = 0.492 m/s²

Next, we find time using the first equation of motion;

V = U + at

Where;

V is the final velocity.

U is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Making time, t the subject of formula, we have;

t = \frac{V - U}{a}

Substituting into the equation, we have;

t = \frac{29 - 4}{0.492}

t = \frac{25}{0.492}

Time, t = 50.8 seconds.

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Speed = (2.3 m) / (3 sec)

Speed = (2.3/3) (m/s)

<em>Speed = 0.766... m/s</em>

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Assessing how well one variable predicts another variable is to blank as detecting cause-effect relationships between different
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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:  

p_{o}=p_{f} (1)  

Where:  

p_{o}=m_{1} V_{o} + m_{2} U_{o} (2)  

p_{f}=(m_{1} + m_{2}) V_{f} (3)

m_{1}=110 kg is the mass of the first football player

V{o}=-7 m/s is the velocity of the first football player (to the south)

m_{2}=75 kg  is the mass of the second football player

U_{o}=4 m/s is the velocity of the second football player (to the north)

V_{f} is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f} (4)  

Isolating V_{f}:

V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}} (5)

V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg} (6)

V_{f}=-2.54 m/s (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy \Delta K is defined as:

\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2} (8)

Simplifying:

\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2}) (9)

\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2}) (10)

Finally:

\Delta K=-2351.25 J (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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CAN SOME ONE TALK TO ME PLZ ASAP IM GOING THROUGH A BAD TIME
ss7ja [257]

Hey! How's it going? If you need anything, feel free to send me a friend request and message me.

Don't worry if things get wrong, they will surely get better, if not, I'm here to talk to you. :)

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