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vovikov84 [41]
4 years ago
9

Calculate the kinetic energy in joules of a 1500 kg automobile moving at 17 m/s . express your answer to two significant figures

and include the appropriate units.
Physics
1 answer:
Sidana [21]4 years ago
6 0
Kinetic energy = 1/2 (mass)*(velocity)^2

ke = 1/2 (1500kg) (17m/s)^2
ke = 216750 kg*m^2/s^2 = 216750 Joules

(sorry, I don't know significant figures. but this is otherwise the correct answer with the right units. hope this helps!! :))
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Answer:

A. At warm tempetures, molecules move around more.

Explanation:

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A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en
posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. <em>If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.</em>

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. <em>Now</em><em> </em><em>please</em><em> </em><em>see</em><em> </em><em>the</em><em> </em><em>picture</em><em>,</em><em> </em><em>which</em><em> </em><em>shows</em><em> </em><em>the </em><em>relation</em><em> </em><em>between </em><em>the </em><em>J </em><em>and </em><em>other</em><em> SI units</em><em>.</em><em> </em><em>Please</em><em> </em><em>understand</em><em> </em><em>that</em><em> </em><em>you</em><em> </em><em>can</em><em> </em><em>construct</em><em> </em><em>your</em><em> </em><em>'own'</em><em> </em><em>formulas</em><em> </em><em>based</em><em> </em><em>these</em><em> </em><em>units</em><em>.</em><em> </em><em>Now</em><em> </em><em>here</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>time</em><em> </em><em>saver</em><em>:</em>

Because almost always the right units are <em>given</em> in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

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3 0
3 years ago
Two remote control cars with masses of 1.16 kilograms and 1.98 kilograms travel toward each other at speeds of 8.64 meters per s
Black_prince [1.1K]

The initial momentum of the system can be expressed as,

p_i=m_1u_1+m_{2_{}}u_2

The final momentum of the system can be given as,

p_f=m_1v_1+m_{2_{}}v_2

According to conservation of momentum,

p_i=p_f

Plug in the known expressions,

\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}

Initially, the second mass move towards the first mass therefore the initial speed of second mass will be taken as negative and the recoil velocity of first mass is also taken as negative.

Plug in the known values,

\begin{gathered} v_2=\frac{(1.16\text{ kg)(8.64 m/s)+(1.98 kg)(-3.34 m/s)-(1.16 kg)(-2.16 m/s)}}{1.98\text{ kg}} \\ =\frac{10.02\text{ kgm/s-}6.61\text{ kgm/s+}2.51\text{ kgm/s}}{1.98\text{ kg}} \\ =\frac{5.92\text{ kgm/s}}{1.98\text{ kg}} \\ \approx2.99\text{ m/s} \end{gathered}

Thus, the final velocity of second mass is 2.99 m/s.

3 0
1 year ago
A person fires a 38 gram bullet straight up into the air. It rises, then falls straight back down, striking the ground with a sp
Y_Kistochka [10]

Answer:

Force exerted = 25.41 kN

Explanation:

We have equation of motion

      v² = u²+2as

u = 345 m/s, s = 8.9 cm = 0.089 m, v = 0 m/s

     0² = 345²+2 x a x 0.089

       a = -668679.78 m/s²

Force exerted = Mass x Acceleration

Mass of bullet = 38 g = 0.038 kg

Acceleration = 668679.78 m/s²

Force exerted = 25409.83 N = 25.41 kN

4 0
4 years ago
An 8.75 kg point mass and a 14.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be
marysya [2.9K]

Answer

given,

mass of the = m₁ = 8.75 Kg

another mass of the object = m₂ = 14 Kg

distance between them = 50 cm

R₁ = 17 cm

R₂ = 50 -17 = 33 cm

a) Force applied due to the Mass 8.75  in +ve x- direction

F_1 = \dfrac{GM_1 m}{R_1^2}

F_1 = \dfrac{6.67\times 10^{-11} \times 8.75\ m}{0.17^2}

F_1 = 2.019\times 10^{8}\ m

Force applied due to mass 14 Kg in -ve x-direction

F_2 = \dfrac{GM_2 m}{R_2^2}

F_2 = \dfrac{6.67\times 10^{-11} \times 14\ m}{0.33^2}

F_2 = 0.857\times 10^{8}\ m

net force

F = F₁ + F₂

F = 2.019\times 10^{8}\ m - 0.857\times 10^{8}\ m

F = 1.162\times 10^{8}\ m

Using newton second law

a = \dfrac{F}{m}

a = \dfrac{ 1.162\times 10^{8}\ m}{m}

a =1.162\times 10^{8} \ m/s^2

b) As the acceleration of mass comes out to be  +ve hence, the direction will be toward the mass of 8.75 Kg

6 0
3 years ago
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