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miv72 [106K]
3 years ago
15

Why is speed greater at the end of a ramp than it is at the beginning of a ramp?

Physics
1 answer:
UkoKoshka [18]3 years ago
5 0
At the begging nothing is happening, but when you push it, it gains speed that is why the end of a ramp is faster.
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3. a) If the distance between two objects increases by a factor of 5 (5),
irina [24]

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

3 0
2 years ago
Use the information to answer the following question.
Nostrana [21]

Answer:

The answer is B. :)

Explanation:

5 0
3 years ago
Middle school physics ​
Alecsey [184]

Answer:

8 V

Explanation:

There is no resistance between the left legs of voltmeters 2 and 3 and there is no resistance between the right legs of voltmeters 2 and 3. They are measuring the same voltage.  

5 0
3 years ago
Which of the following statements is TRUE about impulse? (Consider the Impulse-Momentum Theorem: F▲t = m ▲v )
kotykmax [81]

The statement “Impulse is a vector quantity” is true about Impulse.

Answer: Option B

<u>Explanation: </u>

The object’s action by applied force in a particular time interval, there happens changing in momentum called impulse. It is denoted by a symbol ‘J’ or ‘imp’ and expressed in a unit ‘Ns’. As impulse depends on the acted force, when a collision arises from front, behind or side, the force’s direction would be differed.

                   \text {Impulse }=\text {Force } \times \text {time}=\vec{F} \Delta t

So, from this option A is false as impulse is not a force but changing momentum. The unit is not Newton, it is Newton second (Ns). The force direction differs (impulse direction) for each cases of collision, so option D also false. Hence, option B seems to be correct. Vector quantity deals with both direction and magnitude and important in motion study.

8 0
3 years ago
Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from
Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

  • Density of the ice, ρ = 917 kg/m³
  • Initial pressure, P₁ = 1.0 bar
  • Final pressure, P₂ = 2.0 bar
  • Temperature, T = - 10 C
  • Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

\Delta G = V(P_2-P_1) \\\\

where;

V is the volume of the ice

Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L

Change in pressure;

P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm

The change in the Gibb's free energy per mole (G);

\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G =  0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J

Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.

Learn more about Gibb's free energy here: brainly.com/question/10012881

3 0
3 years ago
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