A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.

An Airbus A320 jetliner has a takeoff mass of 75,000 kg. It reaches its takeoff speed of 82 m/s (180 mph) in 35 s. What is the t

timofeeve [1]

Answer:

175.5 kN

Explanation:

We find the acceleration of the Airbus A320 jetliner from,

a = (v - u)/t where u = initial velocity of jetliner = 0 m/s (since it starts from rest), v = final velocity of jetliner = 82 m/s and t = time for velocity change = 35 s

So a = (v - u)/t = (82 m/s - 0 m/s)/35 s = 82 m/s ÷ 35 s = 2.34 m/s²

Now, the thrust of the engines on the jetliner T = ma where m = mass of jetliner = 75,000 kg and a = acceleration of jetliner = 2.34 m/s²

T = ma

= 75,000 kg × 2.34 m/s²

= 175500 N

= 175.5 kN

8 0

3 years ago

0.001225 kg/L x 720 000 000L =?

I am Lyosha [343]

Answer:

0.001225 kg/L × 720 000 000 L = 882000 kg

Explanation:

Given:

The equation to solve is given as:

0.001225 kg/L × 720 000 000 L = ?

Let us write each term of the product in terms of power of 10.

As 0.001225 has 6 digits after the decimal place, therefore, we use the exponent 6 for 10 and the sign is negative. This gives,

$0.001225\ kg/L = 1225\times 10^{-6}\ kg/L$

Now, for 720000000 L there are 6 zeros after 720. So, we use exponent 6 but with a positive sign. This gives,

$720000000\ L=720\times 10^{6}\ L$

Now, finding the product, we get:

$0.001225\ kg/L\times 720000000\ L\\\\=1225\times 10^{-6}\ kg/L\times 720\times 10^6\ L\\\\=(1225\times 720)\times (10^6\times 10^{-6})\ (\frac{kg}{L}\times L)\\\\=882000\times 10^{6-6}\ kg\\\\=882000\times 10^0\ kg\\\\=882000\times 1\ kg\\\\=882000\ kg$

Therefore, the product is equal to:

0.001225 kg/L × 720 000 000 L = 882000 kg

6 0

4 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

If you add together all of the forces exerted on a object and get a non zero value that is called

True [87]

Hello There!

If you add together all of the forces exerted on a object and get a non zero value that is called the "NET FORCE" of the object

3 0

3 years ago

Read 2 more answers

Beginning at time t = 0, a student exerts a horizontal force on a box of mass 30 kg, causing it to move at 1.2 m/s toward an ele

Alexandra [31]

Since the initial and the final velocity are just equal, it is implied that the acceleration is zero. This means that the net force acting on the body is also zero. The horizontal force should be equal to the force of friction. The force of friction is equal to the product of the coefficient and the normal force.

Ff = (coefficient of friction) x Fn

The normal force is equal to the object's weight if the surface is horizontal.

Ff = (0.20) x (30 kg) x (9.8 m/s²) = 58.8 N


Thus, the horizontal force exerted must be 58.8 N.

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6 0

3 years ago