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ehidna [41]
3 years ago
13

**100 POINTS**

Physics
1 answer:
lawyer [7]3 years ago
6 0

Answer:

B

Explanation:

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A line _____ on a typical Speed vs. Time graph means an object is experiencing a constant acceleration.
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Answer:

The answer should be C. slanted upward to the right.

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A constant horizontal force F is applied to a 5 kg box on a frictionless horizontal surface. The box starts from rest and moves
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A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
lana66690 [7]

Answer:

8.8 kN

Explanation:

V = 2 m³, W = 40 kN, SG = 1.59

Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN

So the weight becomes 40 - 31.2 = 8.8 kN

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3 years ago
The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge
lesya [120]

Answer:

Loss, \Delta E=-10.63\ J

Explanation:

Given that,

Mass of particle 1, m_1=m =0.66\ kg

Mass of particle 2, m_2=7.4m =4.884\ kg

Speed of particle 1, v_1=2v_o=2\times 6=12\ m/s

Speed of particle 2, v_2=v_o=6\ m/s

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

m_1v_1+m_2v_2=(m_1+m_2)V

V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}

V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}

V = 6.71 m/s

Initial kinetic energy before the collision,

K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)

K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)

K_i=135.43\ J

Final kinetic energy after the collision,

K_f=\dfrac{1}{2}(m_1+m_2)V^2

K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2

K_f=124.80\ J

Lost in kinetic energy,

\Delta K=K_f-K_i

\Delta K=124.80-135.43

\Delta E=-10.63\ J

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

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Aktiviti ( 6<br>b) Bina ayat menggunakan penjodoh bilangan berdasarkan<br>situasi di bawah.<br><br>​
Anna007 [38]

Answer:

mananyeeeee situasiii

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