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3 years ago

QUESTION 1

Physics

2 answers:

mario62 [17]3 years ago

7 0

It's D. If it absorbed it would be turning to steam. I am taking honors chem in high school we are learning this.

LekaFEV [45]3 years ago

6 0

The First Question: The water releases energy which causes the water molecules to have less kinetic and potential energy, changing their configuration from liquid to solid.

The Second: The ice absorbs energy which causes the water molecules to have more kinetic and potential energy, changing their configuration from a solid to a liquid.

Took the test and both were correct.

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An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.

Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

T(u) is tension in upper rope
T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0

2 years ago

A Neglecting air resistance, a ball projected straight upward so it remains in the air for 10 seconds needs an initial speed of

ololo11 [35]

Answer:

The initial velocity is 50 m/s.

(C) is correct option.

Explanation:

Given that,

Time = 10 sec

For first half,

We need to calculate the height

Using equation of motion

$v^2=u^2+2gh$

$h =\dfrac{v^2}{2g}$ ....(I)

For second half,

We need to calculate the time

Using equation of motion

$h =ut+\dfrac{1}{2}gt_{2}^2$

$h=0+\dfrac{1}{2}gt_{2}^2$

$t_{2}=\sqrt{\dfrac{2h}{g}}$

Put the value of h from equation (I)

$t_{2}=\sqrt{\dfrac{2\times v^2}{g^2}}$

$t_{2}=\dfrac{v}{g}$

According to question,

$t_{1}+t_{2}=10$

$t_{1}=t_{2}$

Put the value of t₁ and t₂

$\dfrac{v}{g}+\dfrac{v}{g}=10$

$\dfrac{2v}{g}=10$

$v=\dfrac{10\times g}{2}$

Here, g = 10

The initial velocity is

$v=\dfrac{10\times10}{2}$

$v=50\ m/s$

Hence, The initial velocity is 50 m/s.

3 0

3 years ago

During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal

weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

The Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º

The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

Vectors are decomposed into a coordinate system

The components are added

The resulting vector is constructed

Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

v = $\frac{\Delta d}{t}$

Δd = v t

Where v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

t₁ = 2 min = 120 s

Δd₁ = v₁ t₁

Δd₁ = 600 120

Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s

time t₂ = 1 min = 60 s

Δd₂ = v₂ t₂

Δd₂ = 400 60

Δd₂ = 24 103 m

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

cos (180 -35) = $\frac{x_1}{\Delta d_1}$

sin 145 = $\frac{y_1}{\Delta d1}$

x₁ = Δd₁ cos 125

y₁ = Δd₁ sin 125

x₁ = 72 103 are 145 = -58.98 103 m

y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

cos (90+ 30) = $\frac{x_2}{\Delta d_2}$

sin (120) = $\frac{y_2}{\Delta d_2}$

x₂ = Δd₂ cos 120

y₂ = Δd₂ sin 120

x₂ = 24 103 cos 120 = -12 10³ m

y₂ = 24 103 sin 120 = 20,78 10³ m

The component of the resultant vector are

Rₓ = x₁ + x₂

R_y = y₁ + y₂

Rx = - (58.98 + 12) 10³ = -70.98 10³ m

Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

R = $\sqrt{R_x^2 +R_y^2}$

R = $\sqrt{70.98^2 + 62.08^2}$ 10³

R = 94.30 10³ m

We use trigonometry for the angle

tan θ ’= $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{R_y}{R_x}$

θ '= tan⁻¹ $\frac{62.08}{70.98}$

θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

θ = 180 - θ'

θ = 180 -41.2

θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

v = $\frac{\Delta R}{t_3}$