ILL GIVE BRAINLIEST IF UR CORRECT

When did diesel locomotives replace steam locomotives?

liubo4ka [24]

To keep the energy going every day

7 0

3 years ago

Volcanic eruptions are natural phenomena in the . They sometimes erupt, spilling out lava and dust. The lava solidifies, forming

solniwko [45]

Answer:

B. Geosphere

A. Biosphere

A. Atmosphere

Explanation:

Volcanic eruptions occurs within the Geosphere. The geosphere is the rock solid earth make up of rocks that extends into the deep interior.

Magma formed deep within the crust rises to elevated parts and finally erupts as lava on the surface. When they cool, they solidify to form volcanic rocks.

The volcanic eruptions affects the biosphere significantly. The biosphere is the portion of the earth where all life forms exists.

Gases and ash spewed during an eruption into the atmosphere causes severe changes to weather and leads to pollution. The atmosphere is the gaseous envelope round the earth.

3 0

3 years ago

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20

Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

= 150/90

= 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

= 30/90

= 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

5 0

3 years ago

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

agasfer [191]

$v = 2.45×10^3\:\text{m/s}$

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

$F = \dfrac{d{p}}{d{t}}$ (1)

Assuming that the velocity remains constant then

$F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}$

Solving for $v,$ we get

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)$

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

$F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}$

The exhaust rate dm/dt is

$\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}$

$\;\;\;\;\;= 1.36×10^4\:\text{kg/s}$

Therefore, the velocity at which the exhaust gases exit the engines is

$v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}$

$\;\;\;= 2.45×10^3\:\text{m/s}$

6 0

2 years ago

Ch 27 HW Exercise 27.12 10 of 20 Constants A horizontal rectangular surface has dimensions 2.80 cm by 3.15 cm and is in a unifor

deff fn [24]

Answer:

Magnetic field, B = 0.88 T

Explanation:

It is given that,

The dimension of rectangular surface is 2.80 cm by 3.15 cm. The area of rectangular surface is, $A=8.82\ cm^2=0.000882\ m^2$

Angle between the uniform magnetic field and the horizontal, $\theta=31$

Magnetic flux, $\phi=4\times 10^{-4}\ Wb$

Let B is the magnitude of magnetic field in which the rectangular surface is placed. It is given by :

$\phi=BA\ cos\theta$

$\theta$ is the angle between magnetic field and the area

Here, $\theta=90-31=59^{\circ}$

$B=\dfrac{\phi}{A\ cos\theta}$

$B=\dfrac{4\times 10^{-4}}{0.000882\times cos(59)}$

B = 0.88 T

So, the magnitude of magnetic field is 0.88 T. Hence, this is the required solution.

7 0

3 years ago