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ValentinkaMS [17]

3 years ago

12

Two identical objects in outer space have a head-on collision and stick together. If, before the collision, one had been moving

at 2 m/s and the other at 1 m/s, their combined speed after the collision would be?

1 answer:

julia-pushkina [17]3 years ago

5 0

Answer:

1.5 m/s

Explanation:

Momentum is conserved and conservation of momentum is

p₁ + p₂ = p'₁ + p'₂

or

m₁v₁ + m₂v₂ = m₁v'₁ + m₂v'₂

In our problem, after collision v'₁ will be equal to v'₂.

Since objects are identical m₁ = m₂

m(v₁+ v₂) = 2m x v'₁

(2m/s + 1m/s) = 2v'₁

v'₁ = v'₂ = 1.5 m/s

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Use the drop-down menus to complete the statement about Earth’s magnetism.

Nataly [62]

Answer:

The Earth's magnetism is generated in the core, which is composed of iron that is constantly churning

Explanation:

Magnetic fields are produced by charges in motion, therefore by currents.

The outer core of the Earth consists mainly of melted iron that is in constant motion. This iron in motion actually acts as a giant current, and therefore it is responsible for the creation of the Earth's magnetic field.

The magnetic field of the Earth is very weak, in fact its magnitude is on average between 25 and 65 microtesla (for comparison, normal magnets can even produce magnetic fields of a few millitesla).

However, its role is very important for the Earth: in fact, it provides a shield that blocks most of the harmful radiation coming from the Sun.

5 0

3 years ago

Read 2 more answers

Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ

Leona [35]

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3. $10^{8}$ / m (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . $10^{14}$

And we want to find out the value of λ.

c = λ / T

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . $10^{8}$ / 7.30 . $10^{14}$

λ = 4.12 $10^{-7}$ m

Response: 4.12 $10^{-7}$ m = 412 nm

:-)

6 0

3 years ago

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.60×106 N, one at an angle 13.0 west of north, an

Juliette [100K]

Answer: $W=1.93\times 10^9 J$

Explanation:

Given

Force $F=1.6\times 10^{6} N$

one at an angle of $13^{\circ}$ East of North and another at $13^{\circ}$ West of North

Net Force is in North Direction

$F_{net}=2F\cos 13$

Forces in horizontal direction will cancel out each other

thus Work done will be by north direction forces

$W=2F\cdot \cos 30\cdot s$

here $s=0.7 km$

$W=2\times 1.6\times 10^{6}\cdot \cos 30\cdot 700$

$W=1.93\times 10^9 J$

3 0

3 years ago

A 5.00 g bullet traveling 355 m/s is stopped by lodging in the side of a building. The heat produced is shared between the build

steposvetlana [31]

Answer:

Explanation:

Given

mass of bullet $m=5\ gm$

speed of bullet $v=355\ m/s$

bullet is stopped by building and heat produced is shared between building and bullet

Kinetic Energy of bullet is converted into Thermal energy

Kinetic Energy of bullet $K=\frac{1}{2}mv^2$

$K=0.5\times 5\times 10^{-3}\times (355)^2$

$K=315.06\ J$

So 315.06 J of Energy is converted in to thermal energy

7 0

3 years ago

A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.

tankabanditka [31]

Answer:

$v_o=40.14\ m/s$

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

The horizontal speed is constant and equal to the initial speed $v_o$
The vertical speed is zero at launch time, but increases as the object starts to fall
The height of the object gradually decreases until it hits the ground
The horizontal distance where the object lands is called the range

We have the following formulas

$\displaystyle v_x=v_o$

$\displaystyle x=v_o.t$

$\displaystyle v_y=g.t$

$\displaystyle y=\frac{gt^2}{2}$

Where $v_o$ is the initial horizontal speed, $v_y$ is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

$\displaystyle y=\frac{gt^2}{2}$

Rearranging and solving for t

$\displaystyle 2y=gt^2$

$\displaystyle t^2=\frac{2\ y}{g}$

$\displaystyle t=\sqrt{\frac{2\ y}{g}}$

We then replace this value in

$\displaystyle x=v_o.t$

To get

$\displaystyle v_o=\frac{x}{t}$

$\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}$

$\displaystyle v_o=\sqrt{\frac{g}{2y}}.x$

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing $v_o$

$\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6$

The launch velocity is

$\boxed{v_o=40.14\ m/s}$

7 0

3 years ago