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Pachacha [2.7K]
3 years ago
11

Which statement accurately describes a sample of water during parts a and c of the heating curve

Physics
2 answers:
vivado [14]3 years ago
5 0

Answer:

A and C is about 12 cm away from each other.

Explanation:

Olin [163]3 years ago
3 0

Explanation:

A and C is about 25 ccentimetre away

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the answer your looking for is Optical instrument.

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A boy reaches out of a window and tosses a ball straight up with a speed of 10 m/s. The ball is 20 m above the ground as he rele
kirill115 [55]

Explanation:

It is given that,

Speed of the ball, v = 10 m/s

Initial position of ball above ground, h = 20 m

(a) Let H is the maximum height reached by the ball. It can be calculated using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh'            

h'=\dfrac{v^2}{2g}

h'=\dfrac{(10)^2}{2\times 9.8}

h' = 5.1 m

The maximum height above ground,

H = 5.1 + 20

H = 25.1 meters

So, the maximum height reached by the ball is 25.1 meters.

(b) The ball's speed as it passes the window on its way down is same as the initial speed i.e. 10 m/s.

Hence, this is the required solution.                              

6 0
3 years ago
The two equal strong kids are having a tug a war. What do you expect to happen to the ball in this situation
borishaifa [10]

Answer:as per as Newtons second law, The forces exerted on the rope create tension.

As such,The tension is equal to the applied force.The tension is trasmitted to the opposite side and of the rope delivering the applied force.

Hope this helps.. :)

3 0
3 years ago
An archer shoots an arrow with a velocity of 45.0m/s at an angle of 50.0degrees with the horizontal.An assistant standing on the
garri49 [273]

Answer:

a) u = 30.29 m/s

b) t = 2.09 s

Explanation:

given,

velocity = 45 m/s

angle (θ) = 50°

horizontal velocity = 45 cos 50°

time taken to reach 150 m.

times = \dfrac{150}{45 cos 50^0}

t  =  5.19 s

a) height of arrow

s = u t +\dfrac{1}{2}gt^2

s = v sin \theta \times t+\dfrac{1}{2}gt^2

s = 45 sin 50^0 \times 5.19 -\dfrac{1}{2}\times 9.81\times 5.19^2

s = 46.78 m

v² - u² = 2 g s

u² = 2 × 9.81 × 46.78

u = 30.29 m/s

b) time taken by the apple = \dfrac{u}{g}=\dfrac{30.29}{9.81}

                                             = 3.09 s

time after which it has to be thrown = 5.19-3.09 = 2.1 s

5 0
2 years ago
The crest of one wave with an amplitude of 4 m meets up with the crest of a second
trapecia [35]

Answer:

Constructive

Explanation:

3 0
3 years ago
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