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3 years ago

Fermentation is a _________ process.

Physics

1 answer:

ivanzaharov [21]3 years ago

4 0

Fermentation is an aerobic process which means it uses oxygen. anaerobic is a process with out oxegyn

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2) Why do you think fossil fueled cars are still out numbering alternative and replacement fueled cars even though fossil fuels

Misha Larkins [42]

Answer:

Fossil-fueled cars are still outnumbering the alternative and replacement-fueled cars because they are generally cost-effective and are efficient.

Explanation:

Fossil fuels are non-renewable energy that cannot be replenished. Although many people know how harmful using them can be to the environment, still, people continue to buy cars that are powered by it. This is because, using fossil-fueled cars can help you save money compared to using replacement-fueled cars. Its engines are also more powerful, which means they can arrive at their destination in just a few minutes. This makes it efficient. The cost of maintaining a fossil-fueled car is less expensive compared to replacement-fueled cars. People are naturally keen when it comes to budgeting, thus, many people still buy cars powered by fossil fuels.

3 0

3 years ago

Babies typically say their first words:

deff fn [24]

Answer:

2nd one if you ask me

almost every baby stats to talk after their a year old

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2 years ago

Two particles A and B start simultaneously from a Point P with velocities 20 m/s and 30 m/s respectively. A and B move with acce

zysi [14]

Answer:

20 m/s

Explanation:

Given

u(A) = 20 m/s
u(B) = 30 m/s
acceleration equal in magnitude but opposite in direction

Solving

Velocity of A at Q = 30 m/s
From, P to Q, Δv(A) = 30 - 20 = +10 m/s
Therefore, velocity of B at Q will be decreased by 10 as it is equal in magnitude but opposite in direction to A
Δv(B) = v(B at Q) - u(B at P)
-10 m/s = v(B at Q) - 30 m/s
v(B at Q) = 30 - 10 = 20 m/s

6 0

2 years ago

For an object that travels at a fixed speed along a circular path, the acceleration of the object is

lidiya [134]

For an object that travels at a fixed speed along a circular path, the acceleration of the object is LARGER IN MAGNITUDE THE SMALLER THE RADIUS OF CIRCLE.

8 0

3 years ago

Read 2 more answers

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

2 years ago