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ki77a [65]
2 years ago
10

Frieda stands on a hill with a slope of 62 if her mass is 49 what is the magnitude of the normal force acting on her

Physics
1 answer:
mrs_skeptik [129]2 years ago
8 0
C ? im not sure cs im not that smart but try and see .
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The force of gravitation between two spherical bodies is Gm1
malfutka [58]

Answer:

r is the separation between the two spherical bodies

3 0
2 years ago
Describe how the boiling point of water on top of a mountain would be different from its boiling point at sea level.
Sonbull [250]

Answer:

At elevated altitudes, any cooking that involves boiling or steaming generally requires compensation for lower temperatures because the boiling point of water is lower at higher altitudes due to the decreased atmospheric pressure.

Explanation:

7 0
2 years ago
Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?
elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 ms^{-1}

Frequency = 10 Hz = 10 s^{-1}

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

    v = fλ

Therefore wavelength λ = v/f

                                        = 100 ms^{-1} / 10 ms^{-1}

                                        = 10 m

Hence wavelength = 10 m

4 0
3 years ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change
notsponge [240]
We can solve the problem by using the first law of thermodynamics, which states that:
\Delta U = Q-W
where
\Delta U is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ
6 0
3 years ago
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