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allsm [11]
4 years ago
11

Remove the ice cube tray and the plastic bag from the freezer and let them sit for 3 minutes. Then gently remove the piece of ic

e from the tray and the piece of ice from the plastic bag. Place each piece of ice on a piece of foil. Check the pieces of ice about every three minutes until you can clearly see whether there’s a difference in how fast they are melting. What do you observe? HELPPP HURRYYYY
Physics
1 answer:
Salsk061 [2.6K]4 years ago
8 0

Answer:

That when you leave the ice setting without foil, it melts slower. When you put the piece of ice in the foil it melts faster than just letting it set without foil. I really hope this helps.

Explanation:

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Can someone please help me ​
siniylev [52]

Answer:

M a = (M1 + M2) a = F         Newton's Second Law

F = (M2 - M1) g         net force on the system

a = (M2 - M1) / (M1 + M2) g

a = (9 - 7) / (9 + 7) g = 2 / 16 * 10.0 m/s^2 = 1.25 m/s^2

6 0
2 years ago
Convert 60 km/ hr into m/s
elixir [45]

Answer:

1 KM per minute is the real speed in minutes, turn that into 1000 meters per minute and divided by 60, you get a good number of 16.6666666667 which means you could go 50 meters per 3 seconds

Explanation:

so it would be 16.6666666667 meters per second

4 0
3 years ago
(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t)
dolphi86 [110]

Answer:

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)=\int\limits^a_b {a(t)} \, dt

=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k---------------eqn 1

given v(0)=i

i=c_1i+(-1+c_2)j+(0+c_3)k

c_1=1   c_2=1  c_3=0

from equation 1

V(t)=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k----------eqn 2

now r(t)=\int\limits^a_b {v (t)} \, dt

(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k

given r(0)=j

0i+1j+ok = c_1i+c_2j+(\frac{-1}{4}+c_3)k

c_1=0  c_2=0  c_3 =  \frac{1}{4}

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

8 0
3 years ago
A train starts from a station with a constant acceleration of at = 0.40 m/s2. A passenger arrives at the track time t = 6.0s aft
jok3333 [9.3K]

Answer:

4.8 m/s  

Explanation:

When she catches the train,

  1. They will have travelled the same distance.and
  2. Their speeds will be equal

The formula for the distance covered by the train  is

d = ½at² = ½ × 0.40t² = 0.20t²

The passenger starts running at a constant speed 6 s later, so her formula is

d = v(t - 6.0)

The passenger and the train will have covered the same distance when she has caught it, so

(1) 0.20t² = v(t - 6.0)

The speed of the train is

v = at = 0.40t

The speed of the passenger is v.

(2) 0.40t = v

Substitute (2) into (1)

0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t

Subtract 0.20t² from each side

0.20t² - 2.4t = 0

Factor the quadratic

t(0.20t - 2.4) = 0

Apply the zero-product rule

t =0     0.20t - 2.4 = 0

                   0.20t = 2.4

(3)                      t = 12

We reject t = 0 s.

Substitute (3) into (2)

0.40 × 12 = v

            v = 4.8 m/s

The slowest constant speed at which she can run and catch the train is 4.8 m/s.

A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.

7 0
4 years ago
Given that oxygen- 16 and oxygen -18 both have an atomic nmber of 8, how many electrons, protons, and neutrons do these oxygen a
salantis [7]
Hope this helps you.

3 0
3 years ago
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