All categories

3 years ago

What happens to the demand for butter, if the price of margarine increases? For Economics.

Physics

2 answers:

damaskus [11]3 years ago

8 0

People will buy more butter, and then eventually the margarine would lower and this cycle would keep repeating

Anastaziya [24]3 years ago

7 0

Supply and demand goes up .

You might be interested in

A crate resting on a rough horizontal floor is to be moved horizontally. The coefficient of static friction is 0.36. To start th

luda_lava [24]

Answer:

Explanation:

The direction of force will be in upward direction making an angle of θ with the vertical .

Reaction force R = mg - F cosθ

Friction force = μR

= .36 (mg - F cosθ )

Horizontal component of applied force

= F sinθ

For equilibrium

F sinθ = .36 (mg - F cosθ)

F sinθ + .36 F cosθ =.36 mg

F (sinθ + .36 cosθ) = .36 mg

F R( cosδsinθ +sinδ cosθ) = .36 mg ( Rcosδ = 1 . Rsinδ= .36 )

F R sin( θ+δ ) = . 36 mg

F = .36 mg / Rsin( θ+δ )

For minimum F , sin( θ+δ ) should be maximum

sin( θ+ δ ) = sin 90

θ+ δ = 90

Rsinδ / Rcosδ = .36

δ = 20⁰

θ = 70⁰ Ans

5 0

3 years ago

Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?

antoniya [11.8K]

Answer:

2.When they reach the bottom of the fall

Explanation:

The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).

Thus, the correct option is "2" When they reach the bottom of the fall

7 0

2 years ago

The amount of fluid displaced by a submerged object depends on its

Daniel [21]

The amount of fluid displaced by a submerged object depends on its volume.

7 0

3 years ago

Read 2 more answers

A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.

andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

$m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}$

Energy

$\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})$

$m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}$

Where:

$m_{1}$ , $m_{2}$ - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

$v_{1,o}$ , $v_{2,o}$ - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

$v_{1}$ , $v_{2}$ - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If $m_{1} = 0.400\,kg$ , $m_{2} = 0.900\,kg$ , $v_{1,o} = +5.86\,\frac{m}{s}$ , $v_{2,o} = 0\,\frac{m}{s}$ , the system of equation is simplified as follows:

$2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}$

$13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}$

Let is clear $v_{1,f}$ in first equation:

$0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}$

$v_{1,f} = 5.86-2.25\cdot v_{2,f}$

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

$13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}$

$13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}$

$13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}$

$2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0$

$2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0$

There are two solutions:

$v_{2,f} = 0\,\frac{m}{s}$ or $v_{2,f} = 3.606\,\frac{m}{s}$

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ( $v_{2,f} = 3.606\,\frac{m}{s}$ )

$v_{1,f} = 5.86-2.25\cdot (3.606)$

$v_{1,f} = -2.254\,\frac{m}{s}$

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0

3 years ago

Using the picture above, which ball has the greatest potential energy?

weqwewe [10]

Ball 4 because the higher the elevation is the greater the potential energy it has

4 0

2 years ago