24m+350cm+0.46hm+500mm

Of the 5000 species of mammals, 250 species are carnivorous. What is the ratio of carnivorous to mammals?

Mekhanik [1.2K]

Answer: The ratio of carnivorous species to all mammal species is 1:20

Explanation:

250:5000 = 250/250 = 1

250:5000 = 5000/250 = 20

Answer = 1:20

4 0

2 years ago

Two electrons exert a force of repulsion of 1.2 N on each other. How far apart are they? The elementary charge is 1.602 × 10−19

sergejj [24]

The distance between the charges is 13.86 X 10⁴m

Explanation:

Given:

Force, F = 1.2N

Charge, q₁ = 1.602 X 10⁻¹⁹ C

k = 8.987 X 10⁹ Nm²/C²

Distance, d = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2} \\\\$

Substituting the value in the formula we get:

$1.2 = 8.987X 10^9 X\frac{1.602 X 10^-^1^9 X 1.602 X 10^1^9}{r^2} \\\\1.2 = \frac{23.06 X 10^9}{r^2} \\\\r^2 = 19.22 X 10^9\\\\r = 13.86 X 10^4m$

Therefore, the distance between the charges is 13.86 X 10⁴m

5 0

2 years ago

Which is the best location for storing radioactive wastes?

Lubov Fominskaja [6]

The answer is in chambers carved into the rock of a mountain. Radioactive waste should be well disposed where they are little to no anthropogenic activity that would uncover them. This is especially critical for high-level radioactive waste should be buried deep up to 2000 meters. While low-level radioactive waste can be buried 100 meters into the ground such as in sand, the risk is that it could contaminate the water table.

7 0

3 years ago

Read 2 more answers

Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that

Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

$\mu=\dfrac{T}{V^2}\ kg/m$

$V=\sqrt{\dfrac{T}{\mu}}$

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0

3 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$