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coldgirl [10]
3 years ago
13

) a box weighing 77.0 n rests on a table. a rope tied to the box runs vertically upward over a pulley and a weight is hung from

the other end (fig. 4â45). determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 n, (b) 60.0 n, and (c) 90.0 n.

Physics
1 answer:
sineoko [7]3 years ago
8 0
Refer to the diagram shown below.

T = the tension in the rope
N = the the normal reaction (the force that the table exerts on the box)
W = the hanging weight

Assume that the pulley is frictionless.
For equilibrium,
T = W
and
T + N = 77

Therefore
N = 77 - W

(a) When W = 30 N,
 N = 77 - 30 = 47 n
 Answer: 47 N

(b) When W = 60 N,
 N = 77 - 60 = 17 N
 Answer: 17 N

(c) When W = 90 N
 N = 77 - 90 = - 13 N
There is no normal reaction, and the system is no longer in equilibrium.
Instead, the box will be lifted by a force of 13 N, and the box will accelerate upward.

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3 years ago
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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
Len [333]

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

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8 0
3 years ago
A reconnaissance plane flies 605 km away from
kolezko [41]

Answer:

                      v_{avg}  = 355 m/s  

Explanation:

Distance = 605 km

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Final velocity = v_{f} = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

                        2as = v_{f}^{2}+v_{i}^{2}

Then using first equation of motion, we find time

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Then using the formula of average velocity, we find average velocity

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Second method is very simple

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therefore \frac{3t}{5}\times \frac{1}{4}=45

thus t=300 min

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