explain why an inflated balloon is likely to burst if you leave it near a window with the sun streaming it?

Sati [7]

A motion where gravity is the only force acting upon it

6 0

3 years ago

Cheetah mothers perform a number of different behaviors. They and their cubs stay in one place for only four days, moving on bef

Mekhanik [1.2K]

Answer:

Cheetah cubs are in danger from predators like lions and hyenas which can track their prey by scent and so the mother and her cubs leave an area when their scent is too strong so that they are not hunted and the cubs survive.

Mother Cheetahs also train their cubs to hunt so that they may get food for themselves which will ensure their survival as well thus showing that both of these practices can impact on reproductive success.

4 0

3 years ago

A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and

Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

W = K_f (1)

now let's use conservation of energy

starting point. Highest point A

Em₀ = U = m g h

Final point. Lowest point B

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

mg h = K

to find the height let's use trigonometry

at point A

cos 35 = x / L

x = L cos 35

so at the height is

h = L - L cos 35

h = L (1-cos 35)

we substitute

K = m g L (1 -cos 35)

we substitute in equation 1

W = m g L (1 -cos 35)

let's calculate

W = 0.500 9.8 0.950 (1 - cos 35)

W = 0.842 J

7 0

3 years ago

Tudy the images about geologic time.

Fittoniya [83]

Answer:

The first flowering plants appeared in the Mesozoic era, not the Paleozoic era

Explanation:

The Mesozoic era is well known and most famous because of the rule of the dinosaurs which were the dominant animals for most of this are. Also, it is the era in which the mammals appeared, though they lived in the shadows of the dinosaurs and only became dominant after their extinction. Another important evolution that took place and is not mentioned very often is the appearance of the first flowering plants. This was a revolutionary trait for the plants, and it helped them to survive in the changing climate on Earth. Soon this trait enabled this type of plants to spread out significantly and to become one of the most dominant organisms on the planet in the following era.

6 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago