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3 years ago

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a runner runs 2.88 m/s north. she accelerates .350 m/s ^2 at a -52.0 degree angle at the point in the motion where she is runnin

g directly east what is her displacement y

1 answer:

Vitek1552 [10]3 years ago

3 0

Answer:

uh

Explanation:

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Which of the following correctly identifies the relationship between the immune system and the integumentary system (skin, hair,

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F - The skin is the body's first physical barrier, hairs in the nose filter air as it enters the body, pores in the skin release toxins and pathogens

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2 years ago

A 3.0-kg mass and a 5.0-kg mass hang vertically at the opposite ends of a rope that goes over an ideal pulley. If the masses are

irga5000 [103]

Answer:

The time taken will be 0.553 seconds.

Explanation:

We should start off by finding the force exerted by the rope on the 3kg weight in this case.

Weight of 5kg mass = 5 * 9.81 = 49.05 N

Weight of 3kg mass = 3 * 9.81 = 29.43 N

The force acting upward on the 3kg mass will equal the weight of the 5kg mass. Thus the resultant force acting on the 3kg mass is:

Total force = 49.05 - 29.43 = 19.62 N (upwards)

We can now find the acceleration:

F = m * a

19.62 = 3 * a

a = 6.54 m/s^2

We now use the following equation of motion to get the time taken to travel 1 meter:

$s=u*t+\frac{1}{2} (a*t^2)$

$1=0*t+\frac{1}{2} (6.54*t^2)$

t = 0.553 seconds

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3 years ago

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3 years ago

Read 2 more answers

A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.080

deff fn [24]

Answer:

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

$I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}$

now we have

$M = 0.120 kg$

$m_1 = 0.02 kg$

$m_3 = 0.08 kg$

now we have

$I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}$

so we have

$I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162$

$I = 0.02835$

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

$\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}$

$\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)$

$\omega = 4.32 rad/s$

Now speed of 0.08 kg mass when it reaches to bottom point is given as

$v = \omega \frac{L}{2}$

$v = 4.32 (0.45)$

$v = 1.94 m/s$

3 0

3 years ago