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3 years ago

4 applications of elasticity

Physics

1 answer:

svet-max [94.6K]3 years ago

8 0

Answer:

elasticity

1.price elasticity of demand

2.income elasticity of demand

3.cross elasticity of demand

4.elasticity of supply

Explanation:

1. price elasticity of demand is the degree to which the effective desire for something changes as its price changes. In general, people desire things less as those things become more expensive.

2. income elasticity of demand is the responsiveness of the quantity demanded for a good to a change in consumer income. It is measured as the ratio of the percentage change in quantity demanded to the percentage change in income.

3. cross elasticity of demand or cross-price elasticity of demand measures the responsiveness of the quantity demanded for a good to a change in the price of another good, ceteris paribus.

4.price elasticity of supply is a measure used in economics to show the responsiveness, or elasticity, of the quantity supplied of a good or service to a change in its price.

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2 years ago

A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the

BartSMP [9]

Answer:

$r=1.14m$

Explanation:

$\theta$ is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

$F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB$

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

$F_m=F_c\\qvB=F_c(1)$

$F_c$ is the centripetal force and is defined as:

$F_c=m\frac{v^2}{r}$

Here $v$ is the proton's speed and $r$ is the radius of the circular motion. Replacing this in (1) and solving for r:

$qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}$

Recall that 1 J is equal to $6.242*10^{12}MeV$ , so:

$4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J$

We can calculate $v$ from the kinetic energy of the proton:

$K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}$

Finally, we calculate the radius of the proton path:

$r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m$

8 0

3 years ago

I'm completely lost in physics, the Kinematics and dynamics units.. My exam is coming up.. If a biker travelling at 6.4m/s sees

Alexeev081 [22]

D=rt
when biker A catches biker B, the time they've been riding is the same, so
t=t, or d/r=d/r
the rates are 6.4 and 4.7, so
d/6.4=d/4.7
biker B is 34m ahead, so
(d+34)/6.4=d/4.7
multiply both sides by 6.4*4.7:
4.7(d+34)=6.4d
4.7d+=6.4d+159.8
1.7d=159.8
d=94 meters

Another way to think of it is that biker A gains 1.7 meters on B every second (6.4-4.7=1.5), so the time it'll take for him to gain 34 meters is 34/1.7=20 seconds. In that time, biker B travels 4.7*20=94 meters

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2 years ago

4 applications of elasticity​

4 applications of elasticity