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3 years ago

Please help homework due tomorrow,,,

Physics

1 answer:

Ad libitum [116K]3 years ago

8 0

Cody ...

Everything on this page is solved with the SAME formula !

Distance = (speed) x (time) .

Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
but that's too bad. Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.

1). Speed = (distance covered) / (time to cover the distance)

2). Speed = (distance covered) / (time to cover the distance)

3). Distance = (average speed of travel) x (time traveling at that speed)

4). Time to cover the distance = (distance) / (speed)

5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

Calculate the car's speed.
Calculate the sprinter's speed.

... Look at the two speeds.
Decide which one is faster.

... Subtract the slower one from the faster one.
The difference is the answer to "by how much?" .

6). Distance = (speed) x (time spent moving at that speed)

7). Average speed = (TOTAL distance covered)
divided by
(time to cover the TOTAL distance).

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The same force is applied to two skateboards. One rolls across the room and the other moves a few feet and comes to a stop. Wher

igor_vitrenko [27]

The longer you spend reading and thinking about this question,
the more defective it appears.

-- In each case, the amount of work done is determined by the strength
of the force AND by the distance the skateboard rolls while you're still 
applying the force. Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.

-- We know that the forces are equal, but we don't know anything about
how far each one rolled while the force continued. All we know is that
one force must have been removed.

-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.

-- How far did that one roll while you were still pushing it ?

-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?

-- Did each skateboard both roll the same distance while you continued pushing it ?

I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.

8 0

2 years ago

Read 2 more answers

- Name two elements that have the SAME (rounded) atomic mass:

katovenus [111]

Answer:

Magnesium atomic no. = 24,25,26. These are the two elements which have same atomic no

Explanation:

3 0

2 years ago

A 10-meter rope is lying on the floor and has a mass force of 20 N. How much work is required to raise one end of the rope to a

VMariaS [17]

Answer:10N

Explanation: I think

7 0

3 years ago

A halfback on an apparent breakaway for a touchdown is tackled from behind. If the halfback has a mass of 98 kg and was moving a

uranmaximum [27]

Answer:

The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

Explanation:

Given that,

Mass of halfback = 98 kg

Speed of halfback= 4.2 m/s

Mass of corner back = 85 kg

Speed of corner back = 5.5 m/s

We need to calculate their mutual speed immediately after the touchdown-saving tackle

Using conservation of momentum

$m_{h}v_{h}+m_{c}v_{c}=m_{h+c}v_{h+c}$

Where, $m_{h}$ = mass of halfback

$m_{c}$ =mass of corner back

$v_{h}$ = velocity of halfback

$v_{c}$ = velocity of corner back

Put the value into the formula

$98\times4.2+85\times5.5=(98+85)\times v$

$v=\dfrac{98\times4.2+85\times5.5}{98+85}$

$v=4.80\ m/s$

Hence, The mutual speed immediately after the touchdown-saving tackle is 4.80 m/s

3 0

2 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago