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Contact [7]
3 years ago
10

I REALLY NEED HELP PLEASE!!! CAN SOMEONE ANSWER THIS? At 0 ºC, some amount of energy is required to change 1 kg of water from a

solid into a liquid. If you had a 2 kg piece of ice, what effect would this have on the amount of thermal energy required to change the water from a solid to a liquid? A)It would require energy to be removed from the 2 kg piece of ice. The larger piece of ice already has more total energy than the smaller piece of ice, so energy must be removed for the ice to become liquid. B)It would require less energy to change solid water into liquid water because the energy would spread through the ice more quickly and the ice already has a larger total amount of energy because it is larger than a 1 kg piece of ice. C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice. D)It would still require the same amount of energy to change solid water into liquid water because the entire piece of ice would still gain the same amount of energy in each case.
Physics
2 answers:
Nezavi [6.7K]3 years ago
8 0

Answer:

C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice.

Charra [1.4K]3 years ago
4 0

Explanation:

c. it would require more energy to change ........

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Helpppppppp<br> ill give brainliest
Nina [5.8K]

Answer:

I believe the answer is

B. \: 3 \: red, \: 8 \: yellow, \: 1 \: blue

Explanation:

I may be incorrect because I used to do this a long time ago but I believe I am correct

HOPE THIS HELPS!

(:

4 0
3 years ago
A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th
Nadusha1986 [10]

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration g=-9.8 m/s^2 downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

v(t) = v_0 sin \theta +gt

where

v_0 = 30 m/s is the initial velocity of the rock

\theta=53^{\circ} is the angle

t is the time

Requiring v(t)=0, we find the time at which the heigth is maximum:

0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s

The heigth of the rock at time t is given by

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

x(t) = (v_0 cos \theta) t

where

v_0 cos \theta is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

x=(30)(cos 53^{\circ})(2.44)=44.1 m

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

x(t) = (v_0 cos \theta) t

Substituting

v_0 = 30 m/s\\\theta=53^{\circ}

and t = 7.58 s we find:

x=(30)(cos 53^{\circ})(7.58)=136.8 m

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

x(t) = (v_0 cos \theta) t

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

And substituting the different times, we find:

x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m

y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m

x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m

y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m

x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m

y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m

3 0
3 years ago
104470
sergey [27]

Answer:

Grain

Explanation:

I just got it wrong :)

5 0
4 years ago
a motorcyclist starts from rest and accelerates at rate of 4 meters per second squared north. what is the final velocity of the
MArishka [77]

Answer:

80m<em>/</em><em>s</em>

Explanation:

Final velocity is given by

v=u+at

when a motorcyclist starts from rest, initial velocity (u) =0

therefore

v=0+4*20

v=80m/s

that's the answer

8 0
4 years ago
A pool is to be filled with 60 m3 water from a garden hose of 2.5 cm diameter flowing water at 2 m/s. Find the mass flow rate of
stealth61 [152]

Answer:

Time= 6.12*10^4s

mass flow rate m=0.98kg/s

Explanation:

Given

Volume= 60m^3

diamter= 2.5cm= 0.025m

radius= 0.0125m

area A= πr^2

area A= 3.142*0.0125^2

area A= 4.9*10^-4m^2

the velocity of the flow 2m/s

<u>volume flow rate </u>

V=vA

V=2* 4.9*10^-4

V=9.82*10^-4 m^3/s

<u>Time taken to fill the pool</u>

time= volume/volume flow rate

time= 60/9.82*10^-4

time= 6.12*10^4s

<u>Mass flow rate </u>

m= density *volume flow rate

Assuming the density of water to be 997kg/m^3

m= 997*9.82*10^-4

m=0.98kg/s

6 0
3 years ago
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