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Bas_tet [7]
3 years ago
14

More advanced line dances, that requires more intense steps, can be considered

Physics
1 answer:
garri49 [273]3 years ago
3 0

Answer:

Vigorous activity

Explanation:

Vigorous activity

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What is the change in its velocity, v, during this 0.80-s interval?
lyudmila [28]
This link should help you out https://quizlet.com/40330134/biomechanics-questions-flash-cards/
7 0
3 years ago
Which of the substances in the chart has the greatest density?
svlad2 [7]
A.gold is the answer. As it's density is 19.32 grams per cubic centimeter which is lot more than the other substances.
 
4 0
3 years ago
What are the variables in Gay-Lussac's law? pressure and volume pressure, temperature, and volume pressure and temperature volum
Yanka [14]

Answer:

pressure and temperature (assuming volume is constant)

Explanation:

3 0
3 years ago
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A car is moving at a velocity of 25 km/h and increases velocity to 1200 km/h in 2 min. what is the acceleration?
OlgaM077 [116]

Answer:

a = 2.72 [m/s2]

Explanation:

To solve this problem we must use the following kinematics equation:

v_{f} =v_{o} + a*t

where:

Vf = final velocity = 1200 [km/h]

Vo = initial velocity = 25 [km/h]

t = time = 2 [min] = 2/60 = 0.0333 [h]

1200 = 25 + (a*0.0333)

a = 35250.35 [km/h2]

if we convert these units to units of meters per second squared

35250.35[\frac{km}{h^{2} }]*(\frac{1}{3600^{2} })*[\frac{h^{2} }{s^{2} } ]*(\frac{1000}{1} )*[\frac{m}{km} ] = 2.72 [\frac{m}{s^{2} } ]

3 0
3 years ago
(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

3 0
2 years ago
Read 2 more answers
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