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3 years ago

What is the velocity of an object that has been in free fall for 2.5s?

Physics

1 answer:

yan [13]3 years ago

3 0

<h2 /><h2>✏️ Development:</h2><h2 />

✍ We can this formula:

$\huge {\boxed {\sf v = g \cdot t }}$

Looking like this:

$\huge {\text {$ \sf v = 9,8 \cdot 2,5 \Longrightarrow \boxed {\boxed {\bf v= 24,5 }} $ }}$

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Every second, the Sun converts about 600 million tons of hydrogen into 596 million tons of helium. The remaining 4 million tons

Whitepunk [10]

Answer:

Converted to an amount of energy equal to 4 million tons times the speed of light squared. ejected into space in a solar wind.

Explanation:

The 4 million tons of mass is converted to the amount of energy that is equal to 4 million tons times the speed of light squared. This energy moves from the sun with the help of solar winds and received by the planets present in the solar system. This solar energy moves in the form of solar radiation because there is no medium for propagation so that's why we can say that the mass is converted into energy that moves in the form of radiation in discrete packets.

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3 years ago

Which statement is true?

Jet001 [13]

Answer:

A. Speed is a scalar quantity and velocity is a vector quantity.

Explanation:

A scalar quantity is one that consists of only a numerical value.

Speed is a scalar quantity because only the instantaneous value is indicated, for example the speedometer of a car that tells you your speed at the moment but not where you are going or in what direction are you going.

On the other hand, velocity is a vector quantity. Because it is composed of a <u>magnitude and a direction</u>, for example 10m/s to the south is a velocity, and 10m/s is a speed.

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3 years ago

During what stage of engine operation does the piston move upward in the cylinder and force the burned gases out of the cylinder

trasher [3.6K]

*l Take in air and fuel (Intake)
*l Compress (squeeze) the air and fuel (Compression)
*l Ignite and burn the air-and-fuel mixture (Power)
*l Get rid of the burned fuel gases (Exhaust)The Answer is C.Exhaust

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3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

$F=\frac{kq_1q_2}{d^2}$

In this case we have an electron (-e) and a proton (e), so:

$F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N$

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

$F=2.64*10^{-16}N$

3 0

3 years ago

A 2.00 kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m . The spring has

iogann1982 [59]

Answer:

v = 0.41 m/s

Explanation:

In this case, the change in the mechanical energy, is equal to the work done by the fricition force on the block.
At any point, the total mechanical energy is the sum of the kinetic energy plus the elastic potential energy.
So, we can write the following general equation, taking the initial and final values of the energies:

$\Delta K + \Delta U = W_{ffr} (1)$

Since the block and spring start at rest, the change in the kinetic energy is just the final kinetic energy value, Kf.
⇒ Kf = 1/2*m*vf² (2)
The change in the potential energy, can be written as follows:

$\Delta U = U_{f} - U_{o} = \frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (3)$

where k = force constant = 815 N/m

xf = final displacement of the block = 0.01 m (taking as x=0 the position

for the spring at equilibrium)

x₀ = initial displacement of the block = 0.03 m

Regarding the work done by the force of friction, it can be written as follows:

$W_{ffr} = - \mu_{k}* F_{n} * \Delta x (4)$

where μk = coefficient of kinettic friction, Fn = normal force, and Δx =

horizontal displacement.

Since the surface is horizontal, and no acceleration is present in the vertical direction, the normal force must be equal and opposite to the force due to gravity, Fg:
Fn = Fg= m*g (5)
Replacing (5) in (4), and (3) and (4) in (1), and rearranging, we get:

$\frac{1}{2} * m* v^{2} = W_{ffr} - \Delta U = W_{ffr} - (U_{f} -U_{o}) (6)$

$\frac{1}{2} * m* v^{2} = (- \mu_{k}* m*g* \Delta x) -\frac{1}{2} * k * (x_{f} ^{2} - x_{0} ^{2} ) (7)$

Replacing by the values of m, k, g, xf and x₀, in (7) and solving for v, we finally get:

$\frac{1}{2} * 2.00 kg* v^{2} = (-0.4*2.00 kg*9.8m/s2*0.02m) +( (\frac{1}{2} *815 N/m)* (0.03m)^{2} - (0.01m)^{2}) = -0.1568 J + 0.326 J (8)$

$v =\sqrt{(0.326-0.1568} = 0.41 m/s (9)$

7 0

3 years ago