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2 years ago

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Your teacher burns a piece of steel wool in class, demonstrating the chemical property, flammability. You are curious to see wha

t other things can demonstrate chemical properties. Which of these does not demonstrate a chemical change?

1 answer:

LekaFEV [45]2 years ago

3 0

Answer:

I assume by "which" of these you're looking for an example. Water freezing into ice or water, or evaporation, the process of turning from liquid into vapor, would not be chemical changes.

Explanation:

These are physical changes because they do not form a new substance, a chemical change requires a change in the chemical makeup of the substance.

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

A 17 g audio compact disk has a diameter of 12 cm. The disk spins under a laser that reads encoded data. The first track to be r

sladkih [1.3K]

Answer:

0.00066518 Nm

Explanation:

v = Velocity = 1.2 m/s

r = Distance to head = 2.3 cm

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 2.4 s

Angular speed is given by

$\omega=\dfrac{v}{r}\\\Rightarrow \omega=\dfrac{1.2}{0.023}\\\Rightarrow \omega=52.17391\ rad/s$

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{52.17391-0}{2.4}\\\Rightarrow \alpha=21.73912\ rad/s^2$

Torque

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mR^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}0.017\times 0.06^2\times 21.73912\\\Rightarrow \tau=0.00066518\ Nm$

The torque of the motor is 0.00066518 Nm

6 0

2 years ago

A heavy frog and a light frog jump straight up into the air. They push off in such away that they both have the same kinetic ene

Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

The lighter frog is moving faster than the heavier frog

Explanation:

If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

$K(light) = K(heavy) = \frac{1}{2} *ml*vol^{2} = \frac{1}{2}*mh*voh^{2}$

Now, if the only force acting on the frogs is gravity, when they reach to the maximum height, we can apply the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*hmax = vf^{2} -vo^{2} = 2*(-g)*hmax$

When h= hmax, the object comes momentarily to an stop, so vf =0

Solving for hmax:

$hmax =\frac{vo^{2} }{2*g}$

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

5 0

2 years ago

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.

babymother [125]

Answer

given,

mass of the person, m = 50 Kg

length of scaffold = 6 m

mass of scaffold, M= 70 Kg

distance of person standing from one end = 1.5 m

Tension in the vertical rope = ?

now equating all the vertical forces acting in the system.

T₁ + T₂ = m g + M g

T₁ + T₂ = 50 x 9.8 + 70 x 9.8

T₁ + T₂ = 1176...........(1)

system is equilibrium so, the moment along the system will also be zero.

taking moment about rope with tension T₂.

now,

T₁ x 6 - mg x (6-1.5) - M g x 3 = 0

'3 m' is used because the weight of the scaffold pass through center of gravity.

6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3

6 T₁ = 4263

T₁ = 710.5 N

from equation (1)

T₂ = 1176 - 710.5

T₂ = 465.5 N

hence, T₁ = 710.5 N and T₂ = 465.5 N

4 0

2 years ago

The transmutation of a radioactive uranium isotope, 234/92 U, into a radon isotope, 222/86 Rn, involves a series of three nuclea

BaLLatris [955]

Answer:

The answer is 26/98 how i did this is i divided them mulitiplyed well i cant really explain it but im pretty dure its right

Explanation:

3 0

2 years ago

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