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3 years ago

A toy helicopter is flying straight ahead. A fan is blowing air toward the helicopter's right.

Chemistry

2 answers:

e-lub [12.9K]3 years ago

7 0

Answer:

D.) The direction of the helicopter's motion will change

Alex_Xolod [135]3 years ago

4 0

D. the direction of the helicopter's motion will change

Explanation:

When the fan pushes on the helicopter to the right, the direction of the helicopter motion must change as it flies past it.

Since the problem suggest that the fan must push the helicopter towards the right, it will deviate from its course.

According to newton's first law of motion "an object will continue in its state of rest or uniform motion unless acted upon by an external force".
The external force on the helicopter is given by the blowing fan.
This force will change to flight path of the toy helicopter.

learn more:

Newton's law brainly.com/question/11411375

#learnwithBrainly

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If you added 15,000 calories to 2.0 L of water that was at 25.0 degrees C, what temperature would it be at when you finished?

rewona [7]

Answer:

When we finish, the temperature would be 32.5℃

Explanation:

Density of water = mass/volume

So,

Mass of water = Density × Volume

$\\\\$=1.0 \times 2.0 L$\\\\$=1.0 \frac{g}{m L} \times 2000 m L$\\\\$\quad=2000 g$$

$Q=m \times c \times \Delta T$

where

$\Delta T$ = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0 cal/(g℃))

m is the mass of water

plugging in the values

$15000 \mathrm{Cal}=2000 \mathrm{g} \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times \Delta T$

$\\$\Delta T=\frac{15000 \mathrm{cal}}{2000 \mathrm{g} \times \frac{1.0 \mathrm{cal}}{g^{\circ} \mathrm{C}}}$\\\\$\Delta T=7.5^{\circ} \mathrm{C}$$

Final T = ∆T + Initial T

= 7.5℃ + 25℃ = 32.5℃ (Answer).

5 0

3 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$