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nevsk [136]
3 years ago
8

A person supporting Eugene Talmadge for governor would have been MOST likely to belong to which occupation?

Physics
2 answers:
lisov135 [29]3 years ago
7 0

Answer:

B) Farmer

Explanation:

tee hee i just got this question :)))

tatyana61 [14]3 years ago
5 0

Answer:

B) Farmer

Explanation:

USA Test Prep tested. Is correct

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Gravitational energy can be negative?
AlladinOne [14]

answer:

yes

explanation:

At a separation of the surface of Earth (r=6400km) gravity wants pull the test mass closer and closer. ... So the work done by gravity is NEGATIVE. The gravitational potential energy is negative because us trying to do the opposite of what gravity wants needs positive energy.

6 0
3 years ago
2 Points
Mademuasel [1]

According to Newton's Second Law of Motion :

The Force acting on an Object is equal to Product of Mass of the Object and Acceleration produced due to the Force.

:\implies  Force acting = Mass of the Object × Acceleration

Given : Force = 50 newton and Mass of the Object = 10 kg

Substituting the respective values in the Formula, we get :

:\implies  50 N = 10 kg × Acceleration

:\implies \mathsf{Acceleration = \dfrac{50\;N}{10\;kg}}

:\implies Acceleration of the Object = 5 m/s²

4 0
3 years ago
What is an example of radiation? check all that apply
Katen [24]
There are no appropriate examples in the list you provided with your question.

Examples of radiation:

... sunshine to tan your skin
... radio energy to bring you the news
... X-ray to check your teeth
... microwave to heat up the meatloaf
... flashlight to see where you're going
... RF energy to get an MRI of your knee
... infrared radiation from the campfire to warm your tootsies
... UHF radio waves to make a call or check Facebook with your smartphone
4 0
3 years ago
A rigid, insulated tank whose volume is 10 L is initially evacuated. A pinhole leak develops and air from the surroundings at 1
balandron [24]

Answer:

The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

\frac{dU}{dt}= Q-Wm_ih_i-m_eh_e\\\\

W=0\\\\Q=0\\\\m_e=0

From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

temperature:

T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

For point c:

 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

5 0
3 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
3 years ago
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