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Lubov Fominskaja [6]
3 years ago
7

prc-2) A soccer ball accelerates from rest and rolls 6.5m down a hill for 3.1 seconds. It is then stopped by a tree. Find the fi

nal velocity.
Physics
1 answer:
anastassius [24]3 years ago
7 0

Answer:

15?6 and 17 plus 18 equals e b d f g so correct

Explanation:

yws you're welcome and thank for your help with this and I will be there is it ok if I get a and I can

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The kinetic friction force between a 60.0-kg object and a horizontal surface is 50.0 N. If the initial speed of the object is 25
Vikki [24]

Answer:

375 m.

Explanation:

From the question,

Work done by the frictional force = Kinetic energy of the object

F×d = 1/2m(v²-u²)..................... Equation 1

Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.

Make d the subject of the equation.

d = 1/2m(v²-u²)/F.................. Equation 2

Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.

Note: If is negative because it tends to oppose the motion of the object.

Substitute into equation 2

d = 1/2(60)(0²-25²)/-50

d = 30(-625)/-50

d = -18750/-50

d = 375 m.

Hence the it will slide before coming to rest = 375 m

6 0
3 years ago
IF YOU GET GOOD GRADES ON SCIENCE TESTS PLEASE HELP ME!!!!!
Elena-2011 [213]

Answer:

A i think hope this helps

Explanation:

5 0
3 years ago
Read 2 more answers
A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w
Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

8 0
3 years ago
Identify the situations that have an unbalanced force. Check all that apply. A baseball speeds up as it falls through the air. A
viktelen [127]
<span>A baseball speeds up as it falls through the air.
Yes.  Forces on the balloon are unbalanced. 
The balloon is speeding up, so we know that the downward force
of gravity is stronger than the upward force of air resistance.

A soccer ball is at rest on the ground.
No. The ball is not accelerating, so we know that the forces on it
are balanced.
The downward force of gravity on the ball and the upward force
of the ground are equal.


An ice skater glides in a straight line at a constant speed.
No. The skater's speed and direction are not changing, so he is not
accelerating.  That tells us that the forces on him are balanced.

A bumper car hit by another car moves off at an angle.
Yes.  The direction in which the car was moving changed. 
That's acceleration, so we know that the forces on it are unbalanced,
at least at the moment of impact. 

A balloon flies across the room when the air is released.
Yes.  The balloon was not moving. But when the little nozzle was
opened, it started to zip around the room.  So its speed changed.
And, as it goes bloozing around the room, its direction keeps changing too. 
There's a whole lot of acceleration going on, so we know the forces on it
are unbalanced.</span>
5 0
3 years ago
Read 2 more answers
Use the graph for both answers.
Natasha2012 [34]

1. A. 6.00 sec

The graph shows the velocity of an object (y-axis) versus the time (x-axis). In order to find when the magnitude of the velocity reaches 36.00 km/h, we should find the time t (x-coordinate) at which the velocity (y-coordinate) is 36.

By looking at the graph, we see that this occurs when t=6.00 s.


2. A. positive acceleration

In a velocity-time graph like this one, the slope of the curve corresponds to the acceleration of the object. In fact, acceleration is defined as:

a=\frac{\Delta v}{\Delta t}

where \Delta v is the variation of velocity and \Delta t is the variation of time. We see that this quantity corresponds to the slope of the curve in the graph (in fact, \Delta v represents the increment of the y coordinate, while \Delta t represents the increment of the x coordinate). So, a positive slope means a positive acceleration: in this case, the slope is positive, so the acceleration is also positive.


3 0
4 years ago
Read 2 more answers
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