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mars1129 [50]
3 years ago
11

Rank the following letters in order from greatest to least Potential Energy

Physics
1 answer:
gayaneshka [121]3 years ago
3 0

Answer:

a?

Explanation:

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What is the spring constant of a bungee cord that stretches an additional 1.2
ziro4ka [17]

Answer:

A

Explanation:

4 0
3 years ago
Read 2 more answers
Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an
Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed  s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance  d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1    .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 )     ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0
3 years ago
Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees
Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt)  ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1  

70 = Tamb + (100 - Tamb)*e^(-10k)   ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k)   ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)  

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k)   .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0
3 years ago
A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is d
QveST [7]

Answer:

  The target's velocity is about 1320 m/s in the direction 265.7°.

Explanation:

In order for there to be a collision between missile and target, we must have ...

  (target starting position) + (target movement) = (missile movement)

assuming the missile starts from the origin of all measurements. The missile moves 10.2 seconds before impact, so moves a distance of ...

  (10.2 s)(1350 m/s) = 13,770 m

__

We are interested in the target movement, so we can solve for that:

  (target movement) = (missile movement) - (target starting position)

In terms of meters, this is ...

  (target movement) = 13770∠25° - 23500∠55° ≈ 13467.74∠-94.3°

The target covers this distance in the same 10.2 seconds before collision, so its speed is (13467.74 m)/(10.2 s) ≈ 1320.4 m/s.

As a positive angle, the target's direction is ...

  -94.3° +360° = 265.7°

The direction of the target's velocity is 265.7°.

_____

If you're calculating this by hand, there are a couple of ways you can do it. You can convert to rectangular coordinates and back (perhaps least confusing), or you can use the law of cosines to solve the triangle, then translate angles back to the x-y coordinate plane.

Using rectangular coordinates, we have ...

  13770∠25° = 13770(cos(25°), sin(25°)) ≈ (12479.9, 5819.45)

  23500∠55° = 23500(cos(55°), sin(55°)) ≈ (13479.0, 19250.1)

Then the difference is ...

  (12479.9, 5819.45) -(13479.0, 19250.1) ≈ (-999.188, -13430.6)

and the (3rd-quadrant) angle is ...

  target direction = arctan(-13430.6/-999.188) ≈ -94.3° = 265.7°

__

The target's speed is found by dividing the distance it covers by the time it takes.

  √(13430.6² +999.188²)/10.2 ≈ 1320.36 . . . m/s

3 0
3 years ago
3. The car's mass is 400 kg. It moves at a velocity of 20 m/s. Calculate the car's momentum. *
Orlov [11]

Answer:

momentum=mass×velocity

momentum =400kg×20m/s=8000kg.m/s

7 0
3 years ago
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