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3 years ago

At the periphery of a hurricane the air is , and several kilometers above the surface, in the eye, the air is .

1 answer:

6 0

At the periphery of a hurricane the air is sinking, and several kilometers above the surface, in the eye, the air is sinking.

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5. The volume of physical activity attained by an individual who exercises at a level of 7 METs for 35 min · day-1, 4 days · wk-

Airida [17]

D. 980, this is the best answer because 35 x 7 is 980 :)

3 0

3 years ago

A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a

Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

$v_x = \frac{x}{t}$

$v_x = \frac{67}{4.5}$

$v_x = 14.89m/s$

The vertical component of velocity is

$-h = v_y t -\frac{1}{2} gt^2$

Here,

h = Height

g = Gravitational acceleration

t = Time

$v_y$ = Vertical component of velocity

$-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2$

$-1.23= 4.5v_y - 99.225$

$v_y = 21.77m/s$

The direction of the velocity will be given by the tangent of the components, then

$tan\theta = \frac{v_y}{v_x}$

$\theta = tan^{-1} (\frac{21.77}{14.89})$

$\theta = 55.59\°$

The magnitude is given vectorially as,

$|V| = \sqrt{v_x^2+v_y^2}$

$|V| = \sqrt{14.89^2 +21.77^2}$

$|V| = 26.37m/s$

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0

3 years ago

According to newton's second law of motion, how much force will be required to accelerate an object at the same rate if its mass

guajiro [1.7K]

D.

For every action ther is an equal and opposite reaction.

equal in force, opposite in direction

5 0

3 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

True or false: Balanced forces can change an object's direction?

slava [35]

The statement is false. Balanced forces can NOT change the speed OR direction of an object's motion. (See Newton's #1 law of motion.)

4 0

2 years ago

Read 2 more answers

At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.

At the periphery of a hurricane the air is , and several kilometers above the surface, in the eye, the air is .