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Tamiku [17]
3 years ago
8

Displacement vectors of 4 km north, 2km south, 5 km north, and 5km south combine to a total displacement of A 2 km south B 16 km

south C.16 km north D 2km north?
Physics
1 answer:
Phantasy [73]3 years ago
7 0
The north vectors add up as so the south vectors.  Then subtract the two.  For north its 4 + 5 = 9.  South is 2 + 5 = 7.   Then 9-7 = 2km North (D)
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A girl is bouncing on a trampoline where is her gravitational potential energy a maximum and where is her kinetic energy maximum
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Answer:

When you jump down, your kinetic is converted to potential energy of the stretched trampoline. The trampoline's potential energy is converted into kinetic energy, which is transferred to you, making you bounce up. At the top of your jump, all your kinetic energy has been converted into potential energy. Right before you hit the trampoline, all of your potential energy has  been converted back into kinetic energy. As you jump up and down your kinetic energy increases and decrease.

7 0
3 years ago
A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

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3 years ago
A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.
Galina-37 [17]

Answer:

Change in specific internal Energy=250\ \rm Btu/lb

Explanation:

Given:

  • Mass of the gas, m=0.4 lb
  • Initial pressure and volume are p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\
  • Final pressure and temperature are p_1=480\ \rm lbf/in^2
  • Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\

So the final volume of the gas is calculated.

Work in any isotropic is given by w

w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\

According to the first law of thermodynamics we have

Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\

So the Specific Internal Change is given by

\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb

So the specific Change in Internal energy is calculated.

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it has a rocky core so the gravity from that compacts the gases extremly tight

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netineya [11]

Answer:So, the difference between charging by induction and conduction comes down to the contact of the neutral object and the object used to charge it. Conduction requires direct contact, while induction does not.

Explanation:

8 0
3 years ago
Read 2 more answers
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