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3 years ago

What are the atmospheric conditions over Minneapolis,Minnesota? Check all that apply

Physics

2 answers:

Masja [62]3 years ago

3 0

Answer:

the answer is A & C & E just did the lab

Explanation:

nlexa [21]3 years ago

3 0

Answer:

A-- low humidity

C -- warm air temperature

E -- high air pressure

Explanation:

Just took the lab :)

<Jayla>

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A bicyclist is in a 50-km race. She says she had an average velocity of 35.

zlopas [31]

The correct answer to this is (A. Units Only).

It shows that there is a velocity of 35, but the units are missing.

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4 years ago

А______<br> is a unit of measurement for energy. (7 Letters)

a_sh-v [17]

Answer:

Joule or kilowatt/hour

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3 years ago

How much heat is required to raise the temperature of 0.25 kg of water from 20°C to 30°C

pentagon [3]

Answer:

10500 J/kg/*C

Explanation:

Quantity of heat required=mass of substance x specific heat capacity x change in temperature

Quantity of heat required=0.25 x 4200 x [30-20]

Quantity of heat required=0.25 x 4200 x 10

Quantity of heat required=10500 J/kg/*C

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3 years ago

An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro

Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

$\sum F = m_b a$

$m_Bg-T_B = m_Ba$

$T_B = m_Bg-m_Ba$

In the case of mass A,

$\sum F = m_A a$

$T_A = m_Ag-m_Aa$

Making summation of Torques in the Pulley we have to

$\sum\tau = I\alpha$

$T_BR_0-T_AR_0=I\alpha$

$T_B-T_A=I\frac{a}{R^2_0}$

Replacing the values previously found,

$(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}$

$(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}$

$a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}$

$a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}$

$a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}$

Replacing with our values

$a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}$

$a=0.7736m/s^2$

PART B) Ignoring the moment of inertia the acceleration would be given by

$a' =\frac{(m_B-m_A)g}{(m_B+m_A)}$

$a' =\frac{(8-6.8)(9.8)}{(8+6.8)}$

$a' = 0.7945$

Therefore the error would be,

$\%error = \frac{a'-a}{a}*100$

$\%error = \frac{0.7945-0.7736}{0.7736}*100$

$\%error = 2.7%$

8 0

3 years ago

a net force f accelerates a mass m with an acceleration a. if the same net force is applied to mass 5m, then the acceleration wi

diamong [38]

Answer:

F=ma

a =F/m

a =F/5m ms-2,,,,,

3 0

3 years ago