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asambeis [7]
3 years ago
6

Distinguish between directly proportional and inversely proportional with an example of each

Physics
1 answer:
RideAnS [48]3 years ago
4 0

Consider two variables said to be "inversely proportional" to each other. If all other variables are held constant, the magnitude or absolute value of one inversely proportional variable decreases if the other variable increases, while their product (the constant of proportionality k) is always the same.

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A 1500kg car double its speed from 50km/h to 100km/h. By how many times does the kinetic energy from the car's forward motion in
Rashid [163]

Answer:

Explanation:

Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J  

Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J  

But  ,

4 * 1875000 = 7500000

so the KE has increased by 4 times.

8 0
3 years ago
Read 2 more answers
On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3, length 88.8 cm and diameter 2.30 cm fro
vichka [17]

Answer:

w = 28.25 N

Explanation:

To do this, we need to use two expressions.

First, to calculate the weight of any object, we use the 2nd law of newton. In this case, the weight is:

w = m*g  (1)

However we do not have the mass of the rod. We need to calculate that. To calculate the mass, we'll use the expression of density which is:

d = m/V  

From here, we solve for mass:

m = d * V   (2)

Finally, we can know the volume of the rod, because is cylindrical, therefore, the volume of a cylinder is:

V = π * r² * h   (3)

So, in resume, we need to solve for the volume of the rod, then, the mass ans finally the weight. Let's calculate the volume of the rod, converting the units of centimeter to meters, just dividing by 100:

diameter = 2.3 cm ---> radius = 2.3/2 = 1.15 cm -----> 0.0115 m

Length or height = 88.8 cm ----> 0.888 m

Replacing in (3):

V = π * (0.0115)² * 0.888

V = 3.69x10⁻⁴ m

Now, let's use (2) to calculate the mass:

m = 7800 * 3.69x10⁻⁴

m = 2.88 kg

Finally for the weight, we'll use expression (1):

w = 2.88 * 9.81

<h2>w = 28.25 N</h2><h2>And this is the weight of the rod.</h2>
4 0
3 years ago
Un atleta de 70 kg de masa que ha efectuado un salto de altura cae una vez que ha
Allushta [10]

Answer:

a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards

Explanation:

The forces on the athlete are

a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,

therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards

b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles

c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.

3 0
2 years ago
Which energy transformation occurs when an electrolytic cell is in operation.
hodyreva [135]

Answer:

An electrolytic cell converts electrical energy into chemical energy.

Explanation:

7 0
2 years ago
A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

8 0
3 years ago
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