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3 years ago

A missile is launched upward with a speed that is half the escape speed. What height (in radii of Earth) will it reach?

Physics

1 answer:

timofeeve [1]3 years ago

4 0

Answer:

The height (h) will be: $\frac{3}{4}R =h$

Explanation:

The scape speed equation is given by:

$v_{scape}=\sqrt{\frac{2GM}{R}}$

Now, the speed of the missile is

$v_{missile}=\frac{1}{2}v_{scape}$

$v_{scape}=\frac{1}{2}\sqrt{\frac{2GM}{R}}$

Using the conservation of energy, we can find the maximu height of the missile.

$E_{i}=E_{f}$

$\frac{1}{2}mv_{scape}^{2}-mgR =-mgh$

$\frac{1}{2}\frac{2GM}{4R}-gR =-gh$

$\frac{GM}{4R}-gR =gh$

Let's recall that g = GM/R², using the equivalence principle. When R is the radius of the earth and M is the mass of the earth.

$\frac{1}{4}gR-gR =-gh$

$\frac{1}{4}R-R =-h$

Therefore the height (h) will be:

$\frac{3}{4}R =h$

I hope it helps you!

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An air-plane has an effective wing surface area of 17.0 m² that is generating the lift force. In level flight the air speed over

Lady bird [3.3K]

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

5 0

3 years ago

Read 2 more answers

the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce

anastassius [24]

Answer:

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So $24=32+a\times 4$

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

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3 years ago

A ball bearing of radius of 1.5 mm made of iron of density

Serjik [45]

Answer:

$\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}$

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine ( $\sf \eta$ )

Explanation:

$\boxed{ \bold{v = \frac{2}{9} \frac{( {r}^{2} ( \rho - \sigma)g)}{ \eta} }}$

$\sf \implies \eta = \frac{2}{9} \frac{( {r}^{2}( \rho - \sigma)g )}{v}$

Substituting values of r, ρ, σ, v & g in the equation:

$\sf \implies \eta = \frac{2}{9} \frac{( {(0.15)}^{2} \times (7.85 - 1.25) \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times \frac{145.6191}{2.25}$

$\sf \implies \eta = \frac{2}{9} \times 64.7196$

$\sf \implies \eta = 2 \times 7.191$

$\sf \implies \eta = 14.382 \: poise$

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3 years ago

A ball is thrown off the top of a building and lands on the ground below.

natita [175]

Answer:

Mass and velocity.

Explanation:

Kinetic energy <u>is the energy that an object has due to its movement</u>, mathematically it is represented as follows:

$K=\frac{1}{2} mv^2$

where $m$ is the mass of the object, and $v$ is its velocity at a given point in time.

So we can see that to find the kinetic energy just before the ball hits the gound, we need the quantities:

mass of the ball
velocity of the ball before it hits the ground

With the knowledge of these two quantities the kinetic energy of the ball before touching the gound can be determined.

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3 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is: