The hottest climates on Earth are located near the Equator because this region

Si un cuerpo se encuentra en reposo tiene energía cinética?¿por qué?

posledela

Answer:

This is because, when an object which has possessed kinetic energy under its being in motion hits another object, which is stationary the object at rest receives part of the energy inherent in the body which has kinetic energy and expresses the received energy in the form of motion or some forms different from the initial stationary position.

Explanation:

7 0

4 years ago

A peregrine falcon dives at a pigeon. The falcon starts with zero downward velocity and falls with the acceleration of gravity.

Reptile [31]

Answer:

t = 3.94 s

Explanation:

This can be modeled as a case of free fall motion. Because, the falcon is going down with acceleration, that is equal to acceleration due to gravity. To find the time taken by the falcon to intercept the pigeon, we will use second equation of motion for vertical direction:

$h = v_{i}t + \frac{1}{2}gt^2$

where,

h = height = 76 m

vi = initial speed of falcon = 0 m/s

t = time required = ?

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$76\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\t^2 = \frac{(2)(76\ m)}{9.8\ m/s^2}\\\\t = \sqrt{ 15.49\ s^2}\\$

5 0

3 years ago

A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin

Nesterboy [21]

Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

Explanation:

Given that,

Angle = 37°

Velocity of jet airliner = 693 mph

Velocity of wind = 798 mph

We know that,

The new velocity of the aircraft with respect to the ground

$v=v_{a}+v_{w}$

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

$v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}$

Put the value into the formula

$v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}$

$v=1414.3\ mph$

Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

7 0

3 years ago

A person running down the hallway at a velocity of 4.2 m/s comes to a stop in a time of 1.8 seconds to avoid hitting the wall. W

kumpel [21]

Hshshshhshdhdhdhdhddhdhdhdhdhdh

6 0

3 years ago

Read 2 more answers

A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g

alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

↓ Δ ↓

20 g.................50 cm.................40g

38 cm y cm

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

12cm 50 cm 69cm

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↓ Δ ↓

20 g.................50 cm.................40g

38 cm 19 cm

5 0

3 years ago