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2 years ago

Give an example of a scientific hypothesis. Explain how you would test it, and how it could be falsified (proved untrue).

Physics

1 answer:

kaheart [24]2 years ago

8 0

my hypothesis is that If you drop a piece of buttered toast, it will land butter side down.

I tested it by dropping 10 pieces of buttered toast off the table and noted on which side it landed

It could be falsified cause I just made all of this up. In essence, it's like flipping a coin, 50/50 chance so I could say that 5 landed butter up and 5 landed butter down.

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Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.

Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is $t=0.00129 s$

suppose u is the incoming velocity and v is the final velocity after collision

$m=mass\ of\ ball$

Impulse exerted is given by change in momentum of the particle.

Initial momentum $P_i=m\times u$

Final momentum $P_f=m\times v$

Change in momentum $\Delta P=P_i-P_f$

Impulse $J=F_{avg}\cdot t=\Delta P$

$J=F_{avg}\cdot t=m(u-v)$

$F_{avg}=\frac{m(v-u)}{t}$

$F_{avg}=775.2\times m(v-u)\ N$

5 0

3 years ago

Calculate the molecular weight of Aluminium hydroxide

Vedmedyk [2.9K]

Al(OH)3 = 26.98 + [(16×3) + (1.01×3)] = 26.98 + 51.03 = 78.01 and the unit will be g/mol

5 0

2 years ago

What happens to the coefficient of friction when the weight is increased? Why is this?

Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0

3 years ago

Read 2 more answers

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

The table shows the diameters of the planets in our solar system. Assume that a basketball whose diameter is 25

tino4ka555 [31]

Answer:

Mercury:

.85

pea

Venus:

2.1

gumball

Earth:

2.2

gumball

Mars:

1.2

marble

Uranus:

grapefruit

Neptune:

8.6

softball

Explanation:

I have no clue if I'm right but hopefully, I am

3 0

2 years ago