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marshall27 [118]
3 years ago
6

A student placed 700g of water at 28° c in the freezer.after 6 minutes and 15 seconds the water was transformed to ice.

Physics
1 answer:
Maru [420]3 years ago
7 0

Answer:

is this the full questions

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________ is a force acting through distance.
Andreas93 [3]
High school???
No way
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3 years ago
g A change in the initial _____ of a projectile changes the range and maximum height of the projectile.​
docker41 [41]

Answer:

Velocity.

Explanation:

Projectile motion is characterized as the motion that an object undergoes when it is thrown into the air and it is only exposed to acceleration due to gravity.

As per the question, 'any change in the initial velocity of the projectile(object having gravity as the only force) would lead to a change in the range as well as the maximum height of the projectile.' To illustrate numerically:

Horizontal range: As per expression:

R= (u^{2}*sin2θ)/g

the range depending on the square of the initial velocity.

Maximum height: As per expression:

H= (u^{2} * sin^{2}θ )/2g

the maximum distance also depends upon square of the initial velocity.

​

​

​

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3 years ago
Maureen takes notes in class. Wave Interactions
Monica [59]
A. <span>I .................
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3 years ago
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What is the average speed of an aircraft which travels 600m in 10 seconds ?
matrenka [14]

Answer:

60 meters per second

Explanation:

600/10=60

8 0
2 years ago
Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut
Harman [31]

Answer: a) 11.76 m/s  b) 7.056 m

Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

V_{f}=V_{o}+at (1)  

{V_{f}}^{2}={V_{o}}^{2}+2ad (2)  

Where:  

V_{f} Is the final velocity of the object

V_{o}=0 Is the initial velocity of the object (it was dropped)

a=9.8 m/s^{2} is the acceleration due gravity

d is the height of the tower

t=0.02min=1.2 s is the time it takes to the object to reach the ground

b) Begining with (1):

V_{f}=0+at (3)  

V_{f}=at=(9.8 m/s^{2})(1.2 s) (4)  

V_{f}=11.76 m/s (5)  This is the final velocity of the object

a) Substituting (5) in (2):

(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d (6)  

Clearing d:

d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})} (7)  

d=7.056 m (8)  This is the height of the tower

4 0
3 years ago
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