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Damm [24]
3 years ago
7

You want to arrive at your friend's house by 5pm her house is 240 kilometers away if your average speed will be 80km on the trip

when do you need to leave your house in order to get to her house in time
Physics
1 answer:
Annette [7]3 years ago
3 0
You must leave your house by 2:00 pm. 
what you do is you take your total distance and divide that by speed so in this case it is 240km / 80kmh = 3 hours meaning you have to leave the house by 2:00 pm
 
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Answer:

16

Explanation:

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If the moon were 2 times closer to earth than it is now, the gravitational force between earth and the moon would be
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So much brighter and the moon would be so much darker than it is now because the moon is further away from the moon than it is now...
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4 years ago
If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 is the time it takes the object to fall to the ground. 

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


5 0
3 years ago
5) When a substance changes its state, all BUT ONE of these statements is true That is A) Mass remains constant B) Only the form
expeople1 [14]
C) kinetic energy changes. Gas>liquid>solid
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Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

E=Ep+Ek=m*g*h + (1/2)*m*v²=98 J + 125 J= 223 J

Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

Ek=(1/2)*m*v²=12.5 J.

Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

Ep=m*g*h=0.7*9.8*7=48.02 J

The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick 


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