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3 years ago

Create a hypothesis for this testable question: How does price affect the amount of chocolate people buy?

Physics

1 answer:

Wittaler [7]3 years ago

5 0

Answer:

IF the price of chocolate increases, THEN the amount of chocolate people buy decreases

Explanation:

In a scientific investigation, an observation is first made. Based on this observation, a scientific question is then asked. However, a HYPOTHESIS is given next to explain the question asked. A hyothesis is a testable explanation given to solve an observed problem or provide a possible answer to a question.

The hypothesis must be subject to testing via EXPERIMENTATION. A hypothesis usually goes in an IF, THEN format. In this case with the testable question: How does price affect the amount of chocolate people buy?

A hypothesis that can explain this question is: IF the price of chocolate increases, THEN the amount of chocolate people buy decreases.

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What is the mass of 3 m3 of a substance having density 1200 kg/m3

adell [148]

Answer:

3600 kg

Explanation:

From the question,

Density = Mass/Volume

D = M/V.............................. Equation 1

Where D = Density of the substance, M = mass of the substance, V = Volume of the subtance.

Make M the subject of the equation

M = D×V ............................ Equation 2

Given: D = 1200 kg/m³, V = 3 m³.

Substitute these values into equation 2

M = 1200×3

M = 3600 kg.

Hence the mass of the substance is 3600 kg

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3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

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3 years ago

3. A football is kicked with a speed of 35 m/s at an angle of 40°.

jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

$u_y = u sin \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_y = (35)(sin 40)=22.5 m/s$

b) 26.8 m/s

The initial horizontal velocity is given by:

$u_x = u cos \theta$

where

u = 35 m/s is the initial speed

$\theta=40^{\circ}$ is the angle of projection of the ball

Substituting into the equation, we find

$u_x = (35)(cos 40)=26.8 m/s$

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

$v_y = u_y + at$

where

$v_y$ is the vertical velocity at time t

$u_y = 22.5 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, $v_y =0$ ; substituting, we find the time t at which this happens:

$0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s$

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}gt^2$

where

s is the vertical displacement at time t

$u_y = 22.5 m/s$

$g=-9.8 m/s^2$

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

$s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m$

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

$t=2(2.30 s)=4.60 s$

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

$v_x = 26.8 m/s$

Therefore, the horizontal distance travelled during the whole motion is

$d=v_x t = (26.8)(4.60)=123.3 m$

So, the ball lands 123.3 m far from the initial point.

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3 years ago

A snail and an inchworm are in a race. Their race track heads north for a distance of 2 m. If the inchworm comes to the end of t

olga2289 [7]

Well the basic equation for velocity is v=d/t where d is distance and t is time. So v=2m/50s and the answer is v=0.04meter/second.

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3 years ago

If the frequency of q of a trait is 0.4, what is the frequency of p?

lianna [129]

In genetic traits, p and q represent the relative probabilities of the two alleles manifesting. If these two are the only options (ex. a dominant one and a recessive one), then the probabilities of both must sum up to 1. In this case, since we are given that q = 0.4, then p + q = 1, p + 0.4 = 1, and p = 0.6.

4 0

3 years ago

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