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irina1246 [14]
2 years ago
15

Help me please!!!!!!!!!

Physics
1 answer:
beks73 [17]2 years ago
6 0

limiting frictionl force

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How does the temperature of water change when it is heated on a stove top and then begins to boil?
guapka [62]

Answer:

C

Explanation:

First the water heats up to the boiling point ( temp increases)

  then, as it boils it remains at constant temp ( boiling point)

5 0
2 years ago
A mass is placed at the end of a spring. It has starting velocity of V & allowed to oscillate freely. If the mass has a star
LiRa [457]

Answer:

Equation for SHM can be written

V = w A cos w t        where w is the angular frequency and the velocity is a                                         maximum at t = 0

V1 = w1  A cos w1 t

V2 = w2 A cos w2 t

V2 / V1 = w2 / w1     since cos X t = 1 if t = zero

V2 / V1 = 2 pi f2 / (2 pi f1) = f2 / f1 = T1 / T2

If the velocity is twice as large the period will be 1/2 long

8 0
2 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?​
valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2)    (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

5 0
2 years ago
3. The center of mass (or center of gravity) of a two-particle system is at the origin. One particle
nika2105 [10]

Answer:

B) (-2.0 m, 0.0 m)

Explanation:

Given:

Mass of particle 1 is, m_1=2.0\ kg

Mass of particle 2 is, m_2=3.0\ kg

Position of center of mass is, (x_{cm},y_{cm})=(0,0)

Position of particle 1 is, (x_1,y_1)=(3.0\ m,0.0\ m)

Position of particle 2 is, (x_2,y_2)=(?\ m,?\ m)

We know that, the x-coordinate of center of mass of two particles is given as:

x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}

Plug in the values given.

0=\frac{(2.0\times 3)+(3.0\times x_2)}{2.0+3.0}\\\\0=6+3x_2\\\\3x_2=-6\\\\x_2=\frac{-6}{3}=-2.0\ m

We know that, the y-coordinate of center of mass of two particles is given as:

y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}

Plug in the values given.

0=\frac{(2.0\times 0)+(3.0\times y_2)}{2.0+3.0}\\\\0=0+3y_2\\\\3y_2=0\\\\y_2=\frac{0}{3}=0.0\ m

Therefore, the position of particle 2 of mass 3.0 kg is  (-2.0 m, 0.0 m).

So, option (B) is correct.

8 0
3 years ago
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