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AleksandrR [38]
3 years ago
12

A force of 100 N is applied to a crate at an angle of 60° to the horizontal surface. The force causes the crate to move 2 m acro

ss the floor. How much work is done on the crate?
a. 50 J
b. 100 J
c. 200 J
Physics
1 answer:
MAVERICK [17]3 years ago
4 0

Answer: B 100 J

Explanation: Pic Attached. Hope This Helps.

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Answer:

W = 9.6 N

Explanation:

Given that,

Area on 1 foot, A = 0.6 m²

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The pressure is given by force acting per unit area. So,

P=\dfrac{F}{A}\\\\P=\dfrac{W}{A}\\\\W=16\times 0.6\\\\W=9.6\ N

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam
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Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

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

v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s

b) the change in the combined kinetic energy of the two-car system during this collision

\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))

substitute the value in the equation above

=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J

Hence, the change in combine kinetic energy is -2534.78J

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