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Bond [772]
4 years ago
14

Real life example in which linear motion is not uniform motion.

Physics
1 answer:
wlad13 [49]4 years ago
5 0
1. A horse running in a race
2. A car in traffic
3. Movement of an asteroid
4. A car coming to a halt


Hope this helped
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When a boat moves through the water, the waves in front of the boat bunch up, while the waves behind the boat spread out. this i
liberstina [14]

Answer: Doppler Effect

Doppler Effect can be described as the change in wavelength of a wave like upward shift in frequency for an object whom is approaching and an apparent downward shift in frequency for observers from whom the source is receding. This effect can be observed when a boat moves through the water then the waves in front bunch up while the waves behind the boat spread out.

8 0
4 years ago
A 19kg block is being pulled with a constant horizontal force of 95 Newton’s while also experiencing a constant friction force o
yuradex [85]

Answer:

A

⋅(19kg)=19Akg

95=19Akg

19=19Ak

Explanation:

5 0
3 years ago
Linear expansivity?<br>​
shusha [124]

Linear expansivity, area expansivity and volume or cubic expansivity are

7 0
3 years ago
Two circles are drawn in a 12-inch by 14-inch rectangle. Each circle has a diameter of 6 inches. If the circles do not extend be
Dennis_Churaev [7]

Answer:

d = 10 inch

Explanation:

The farthest distance between the centers, is along the diagonal of the rectangle. Therefore, we need to calculate the diagonal of the rectangle, but counting the fact that we have both circles.

So if, one side is 12 inch, and the other is 14 inch, we can use the Pitagoras theorem which is:

d = √(a²) + (b)²

Where a and b, are the lenght of the rectangle, but without the lenght  of the diameter of both circles.

With this, the expression is this:

d = √(14 - 6)² + (12 - 6)²

d = √64+36

d = √100

d = 10 inches

6 0
3 years ago
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
4 years ago
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