Ask question

Ask question

All categories

2 years ago

14

A hedgehog runs 500 meters in 10 minutes, then 300 meters in 8 minutes and finally 100 meters in 6 minutes. What is the hedgehog

's average speed in meters per minutes?

1 answer:

Marina86 [1]2 years ago

5 0

200 meters per minutes

500 meters -300 meters= 200 meters

300 meters -100 meters =200 meters

You might be interested in

A parallel-plate capacitor has area A and plate separation d, and it is charged to voltage V. Use the formulas from the problem

Shtirlitz [24]

Answer:

U = (ε0AV^2) / 2d

Explanation:

Where C= capacitance of the capacitor

ε0= permittivity of free space

A= cross sectional area of plates

d= distance between the plates

V= potential difference

First, the capacitance of a capacitor is obtained by:

C = ε0A/d.

Starting at the formula , U= (CV^2)/2. Formula for energy stored in a capacitor

Substitute in for C:

U = (ε0A/d) * V^2 / 2

Hence:

U = (ε0AV^2) / 2d

3 0

2 years ago

Brainliest if correct

Bad White [126]

Answer:

D: Increase the distance between the objects.

E: Decrease the mass of one of the objects.

6 0

1 year ago

The type of energy that depends on position is called

attashe74 [19]

The type of energy that depends on position is called
kinetic energy

5 0

3 years ago

A 2.5g copper penny is given a charge of -4.0*10^-9c. how mny excess electrons are on the penny?

Lyrx [107]

Answer : The excess of electrons on the penny are, $2.5\times 10^{10}$ electrons

Solution : Given,

Total charge = $-4.0\times 10^{-9}C$

Charge on electron = $-1.6\times 10^{-19}C$

Formula used :

$\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}$

Now put all the given values in this formula, we get the excess of electrons present on the penny.

$\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}=\frac{-4.0\times 10^{-9}C}{-1.6\times 10^{-19}C/e^-}=2.5\times 10^{10}e^-$

Therefore, the excess of electrons on the penny are, $2.5\times 10^{10}$ electrons

7 0

2 years ago

Light passes through a pair of very thin parallel slits. The resulting interference pattern is viewed far from the slits at vari

butalik [34]

Answer:

Intensity of the next bright fringe will remain same.

Explanation:

The question is based on Young's double slit experiment, since its about bright fringe, the interference here is constructive.

Young's condition for constructive interference is given by:

dsin $\theta = n\lambda$

where,

d = slits distance from eachother or width of the slits

$\lambda$ = wavelength

n = interferance order

Also, we know that in Young's experiment, the fringe intensity is given by:

$I' = 4Icos^{2}\phi$

where,

$\phi$ = phase difference

Therefore, in absence of phase difference i.e., $\phi = 0$ , the intensity of the next bright fringe will not change and it will remain same.

4 0

2 years ago