All categories

3 years ago

What are 3 characteristics of a tornado?

Physics

2 answers:

BabaBlast [244]3 years ago

8 0

Answer: shape, size, wind speed

Explanation:

IgorLugansk [536]3 years ago

6 0

Answer:

Explanation:

Features. A tornado is a swirling vortex or column of air with a hollow core. The circulating air often contains debris and dust and moves in an upward spiral at high speeds. The bottom of the tornado column makes contact with the ground, while the top of the tornado can extend 5 or more miles into the sky.

air
shape
speed

You might be interested in

A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s

Katen [24]

It will be traveling exactly 24 miles per hour <span />

7 0

3 years ago

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0

3 years ago

The velocity of a particle moving along the x-axis varies with time according to v(t) = A + Bt−1 , where A = 2 m/s, B = 0.25 m,

kondaur [170]

Answer:

$a= -2\ m/s^2$

$a=-12.5\ m/s^2$

x=2.17 m

x=8.4 m

Explanation:

Given that

$v=A+Bt^{-1}$

$v=2+0.25t^{-1}$

To find acceleration :

we know that

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=0-0.5t^{-2}$

$a=-0.5t^{-2}$

Acceleration at t= 2 s

$a=-0.5\times 2^{-2}$

$a= -2\ m/s^2$

Acceleration at t= 5 s

$a=-0.5\times 5^{-2}$

$a=-12.5\ m/s^2$

We know that

$v=\dfrac{dx}{dt}$

$dx=\left(2+\dfrac{1}{4t}\right)dt$

Position at t= 2 s:

$\int_{0}^{x}dx=\int_{1}^{2} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{2}$

x=2+0.25 ln2

x=2.17 m

Position at t= 5 s:

$\int_{0}^{x}dx=\int_{1}^{5} \left(2+0.25\dfrac{1}{t}\right)dt$

$x=\left [2t+0.25\ lnt \right ]_{1}^{5}$

x=8+0.25 ln5

x=8.4 m

4 0

2 years ago

Explain what is meant by “field” and compare the properties of gravitational, electric, and magnetic forces in terms of particle

notsponge [240]

Field in this context refers to a region of the space to which corresponds a value.

There is a gravitational field around the earth, because a mass m placed at any point around the earth will be atracted (gravitational force) by it.

There is an electric field in a point when a charge placed there feels an electric force.

The gravitational field is proportional to the value of the mass of the object that produces it.

The electric field is proportional to the magnitude of the charge of the particle that produces it.

The gravitational field is always attractive. The electric field may be attractive or repulsive.

Both fields are proportional to the inverse of the squared distance.

The magnetic field is created when a charge is in movement,i.e a charge in movement will create a magnetid fiedl around it that will act and create a magnetic force over other charge also in movement.

The magnetic field is proportional to the product of the charge times its velocity and inversely proportional to the squared distance. The force generated my be attractive or repulsive.

6 0

3 years ago

A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le

xz_007 [3.2K]

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

4 0

2 years ago