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liberstina [14]

3 years ago

12

Two cars (with masses 2000 kg and 1000 kg) collide head on an icy road. Before the collision, the more massive vehicle was movin

g at 10 m/s to the right and the less massive vehicle was moving at 15 m/s to the left. Immediately after the collision, the more massive vehicle was moving at 7.0 m/s to the left. How fast was the less massive Vehicle immediately after the collision?

1 answer:

mezya [45]3 years ago

5 0

Answer:

v₂' = 19 m/s

Explanation:

given,

mass of car one = m₁ = 2000 Kg

mass of car two = m₂ = 1000 Kg

velocity of car one in right (v₁) = 10 m/s

velocity of car two in left (v₂)= 15 m/s

after collision

velocity(v₁') = 7 m/s

v₂' = ?

using conservation of momentum

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

2000 x 10 - 1000 x 15 = - 2000 x 7 + 1000 v₂'

1000 v₂' = 19000

v₂' = 19 m/s

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No one can answer that kind of question since it was a specific project.

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Meteor Infrasound A meteor that explodes in the atmosphere creates infrasound waves that can travel multiple times around the gl

ladessa [460]

Answer:

The frequency of these waves is $4.27\times10^{-2}\ Hz$

Explanation:

Given that,

Wavelength = 6.6 km

Distance = 8810 km

Time t = 8.67 hr

We need to calculate the velocity of sound

Using formula of velocity

$v = \dfrac{D}{T}$

Where, D = distance

T = time

Put the value into the formula

$v =\dfrac{8810}{8.67}$

$v=1016\ km/hr$

We need to calculate the frequency

Using formula of frequency

$v=n\lambda$

$n=\dfrac{v}{\lambda}$

Put the value into the formula

$n=\dfrac{1016}{6.6}$

$n=153.93\ hr$

$n=\dfrac{153.93}{60\times60}$

$n=0.0427\ Hz$

$n=4.27\times10^{-2}\ Hz$

Hence, The frequency of these waves is $4.27\times10^{-2}\ Hz$

8 0

3 years ago

What name is given to the condition of stability enabled by a set of processes such as those involved in the maintenance of inte

BabaBlast [244]

Answer:

C. Metabolism

Explanation:

Although it is a process that we cannot visualize, we feel it daily in our day by day life. A clear example is when we consume food of any kind, each of these foods contains a reference amount of energy in kilocalories (kcal). These foods supply energy to the human body, this energy from food is used for different day actions, and much to maintain body temperature. When the human body is exposed to very low outside temperatures, the body is exposed to a considerable loss of energy in the form of heat transfer, so the person will begin to feel hungry so that the body needs calories from food to maintain body temperature at stable levels. Metabolism is essential in the human body, at the moment when the human being dies his temperature decreases because the metabolism stops working.

3 0

2 years ago

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the

LiRa [457]

Answer:

a) M = 2,939 10¹⁷ kg , b) $M_{e}$ / M = 2 10⁷

Explanation:

a) The equation for gravitational force is

F = G m M / r²

Where G is the gravitational constant that is worth 6.67 10⁻¹¹ N m² / kg², m the mass epa person. M the mass of the Mountain and r the distance between them.

The value of this force is 2% of the person's weight

F = 0.02 W = 0.02 mg

we replace

0.02 mg = G m M / r²

M = 0.02 g r² / G

r = 10 km = 10 10³ m = 1.0 10⁴ m

M = 0.02 9.8 (10⁴)² / 6.67 10⁻¹¹

M = 2,939 10¹⁷ kg

b) to compare the masses we find their relationship

$M_{e}$ / M = 5.98 1024 / 2,939 1017

$M_{e}$ / M = 2 10⁷

c) treating the mountain as a point object

d) The mountain is not spherical so the distance changes depending on the height of the mountain

8 0

3 years ago

A plane flying horizontally at an altitude of 1 mi and a speed of 560 mi/h passes directly over a radar station. Find the rate a

natulia [17]

Given:

altitude, x = 1 mile

speed, v = 560 mi/h

distance from the station, x = 4 mi

Solution:

To find the rate,

$\frac{dx}{dt} = 0$

Now, from the right angle triangle in fig 1.

Applying pythagoras theorem:

$h^{2}=x^{2} + y^{2}$

differentiating the above eqn w.r.t 't' :

$2h\frac{dh}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ (1)

Now, putting values in eqn (1):

$2h\frac{dh}{dt} = 2\times 1\times 0 + 2y\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{y}{h}\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}$

$\frac{dh}{dt} = \frac{560}{4}\frac{dy}{dt}$

$\frac{dh}{dt} = 140\sqrt{4^2 - 1}$

The rate at which distance from plane to station is increasing is:

$\frac{dh}{dt} = 542.22 mph$

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3 years ago