All categories

2 years ago

A meteorologist tries to forecast the weather by comparing the past and _______ areas on a weather map.

Physics

1 answer:

Gnom [1K]2 years ago

7 0

Answer:

The answer is D current

Explanation:

PLZ give brainliest if this helped

You might be interested in

If the current density in a wire or radius R is given by J-k+5,0F wire? R, what is the current in the wire?

Deffense [45]

Answer:

I = $R^{2}$ (K+5)

Explanation:

Given :

J = k+5

Now selecting a thin ring in the wire of radius "r" and thickness dr.

Current through the thin ring is

dI = J X 2πrdr

dI = (K+5) x 2πrdr

Now integrating we get

I = $\int_{0}^{R} = (K+5).2\pi rdr$

I = (K+5) 2π $\int_{0}^{R} rdr$

I = (K+5) 2π $\frac{R^{2}}{2}$

I = $R^{2}$ (K+5)

8 0

3 years ago

The graphic organizer above shows that the properties of waves are influenced by the energy of waves. Name 2 properties of waves

Stells [14]

Amplitude: the height of the wave<span>, measured in meters
</span><span>Wavelength: the distance between adjacent crests, measured in meters
</span>

3 0

2 years ago

One 10.0 Ω resistor is wired in series with two 10.0 Ω resistors in parallel. A 45.0 V battery supplies power to the circuit. Wh

RideAnS [48]

The answer you are looking for 3.0 A

4 0

2 years ago

Read 2 more answers

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

2 years ago

A 0.45kg baseball is pitched towards home plate at 20 m/s. The ball is hit back towards the pitcher with a speed of 30 m/s. What

Inessa05 [86]

Answer:

4.5kgm/s

Explanation:

Change in momentum is expressed as

Change in momentum = m(v-u)

M is the mass

V is the final velocity

u is the initial velocity

Given

m=0.45kg

v = 30m/s

u = 20m/s

Substitute

Change in momentum = 0.45(30-20)

Change in momentum = 0.45×10

Change in momentum = 4.5kgm/s

3 0

2 years ago