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3 years ago

True or false Are African Grey birds smart .

Physics

2 answers:

Pavel [41]3 years ago

5 0

Answer:

African grey birds are not only really smart, they are helpful too. They are the first bird species to pass a test that requires them both to understand when another animal needs help and to actually give assistance.

which i think is pretty awesome and when they meet a person they start the boding process and their really sweet and kind at first but anyways

its true

notka56 [123]3 years ago

3 0

This is true they are very smart.

You might be interested in

What is the difference between weathering, erosion, and decomposition?

ladessa [460]

Weathering is where something is changed or mutilated by the weather. Erosion is hwere water or an other substance digs awaya at something and wears it out. I don't exactly know the last one or how to explain

8 0

3 years ago

Read 2 more answers

In a domestic electric circuit (220.V),an electric kettle of 3kW power rating is operatedthat has a current rating of 4A. What r

kozerog [31]

Answer:

(i) It will take longer (75/22 times) the time to boil a given quantity of water compared to when the current is working at its rated capacity

(ii) The reasons are;

(1) For economy; most of the appliances in the home only require low power circuitry with thinner wire while a separate high power circuitry is created directly from the main supply for the high electric power rated appliances

(2) For safety; to prevent the over heating of the electric circuits when an high electric power appliance needs to be connected an high power electric power outlet has to be specified

Explanation:

(i) The power rating of the kettle = 3 kW

The voltage rating of the circuit, V = 220 V

The current rating, I = 4 A

The formula for electric power = I² × R = I × V

Therefore, we have;

Power produced = 220 V × 4 A = 880 V·A = 800 W

Hence, since the power produced is below the power rating of the electric kettle, it will take a longer time to boil a given amount of water than specified by the kettle manufacturer

The energy supplied H = V×I×t

Where:

t = Time in seconds

Therefore, we have;

3 kW = 3000 W;

3000 × t₁ = 880 × t₂

t₂/t₁ =3000/880 = 75/22

Hence the kettle will take 75/22 multiplied by the time it takes when working at rating capacity to boil a given quantity of water

(ii) This is so because the power consumption already factored in the electrical installation as well as the type of appliances utilized in the home allow for several low power rating consumption and few high power rating consumption

Therefore, for both economy and safety the electrical circuit are split to allow for the use of very thick copper or aluminium electric cables in the high power rating electric circuits to which can be plugged high electric power consuming devices such as the water heater and electric cooker

The low electric power consuming devices, such as the electric bulb and fans are connected to the low or "regular" power rating electric circuit outlets

The current required for high power and low power appliances is different and also the fuse rating required for both the appliances different, so two separate circuits are used for high power and low power appliances.

6 0

3 years ago

Consider a system two point charges. One has charge +q at (x, y,z) -(a,0,0) and another of charge-q at (x, y, z) = (-a, 0,0). 5.

olga2289 [7]

Answer:

electricfield at (0,0,0) is Et = 2 k q / a²

Explanation:

For the first part see the diagram , the field lines start from the positive charge and reach the negative charge, notice that no line should cross, some lines go to infinity

For the second part we use that the electric field is a vector quantity and therefore we add the field of each charge, using the equation

E = k q / r²

Point (0,0,0)

We calculate for the charge -q which is at a distance R = a

E1 = k (-q) / a²

E1 = - kq / a²

As the test charge is positive in the field it goes to the left, attractive force

We calculate for the charge that is also at R = a

E2 = k q / a²

This field goes to the left, repulsive force

We find the total electric field

Et = E1 + E2

Et = kq / a² + kq / a²

Et = 2 k q / a²

Point (0,0, R)

We use the same equations, but with another distance, for the charge -q the distance is R = R+a and for the charge + q the distance is R = R-a

E1 = k q / (R + a)²

E2 = kq / (R-a)²

Et = kq [1 / (R + a)² + 1 / (R-a)²]

Et= kq {[(R-a)² + (R + a)²] / [(R + a)² (R-a)²]}

Et= kq {2 (R² + a²) / [(R + a)² (R-a)²]}

If we use the condition that R> a we can despise in the patents "a"

(R² + a²) = R² (1+ a² / R²) ≈ R²

(R + a)² = R² (1 + a / R)² ≈ R²

(R- a)² = R² (1-a / R)² ≈ R²

Substituting in the total electric field

Et = kq {2 R²) / [R²R²]}

Et =kq 2 / R²

7 0

3 years ago

Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ

bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c, s= 400 m

r= ?

we have to solve for the radius

we know that circumference

$c= 2\pi r$

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular displcament

∅= 400/63.63

∅= 6.28 radians

8 0

2 years ago

A string is wrapped around a pulley with a radius of 2.0 cm. The pulley is initially at rest. A constant force of 50 N is applie

Ray Of Light [21]

Answer:

0.20kg-m^2

Explanation:

Let the linear velocity of the rope(=of pulley) is v m/s

Using kinematic equation

=> v = u + at

=>v = 0 + 4.9a

=>v = 4.9a ------------ eq1

By v^2 = u^2 + 2as

=>v^2 = 0 + 2 x v/4.9 x 1.2

=>4.9v^2 - 2.4v = 0

=>v(4.9v - 2.4) = 0

=>v = 2.4/4.9 = 0.49 m/s

Thus by v = r x omega

=>omega = v/r = 0.49/0.02 = 24.49 rad/sec

BY W = F x s = 50 x 1.2 = 60 J

=>KE(rotational) = W = 1/2 x I x omega^2

=>60 = 1/2 x I x (24.49)^2

=>I = 0.20 kg-m^2

5 0

3 years ago