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Degger [83]
3 years ago
12

A 10000 kg rocket blasts off vertically from the launch pad with a constant upward of 2.25 m/s2 and feels no appreciable air res

istance. When it has reached a height of 525 m, its engine suddenly fails so that only gravity acts on it after that. What is the maximum height that the rocket reaches?
Physics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

The maximum height that the rocket reaches is 645.5 m.

Explanation:

Given that,

Mass = 10000 kg

Acceleration = 2.25 m/s²

Distance = 525 m

We need to calculate the velocity

Using equation of motion

v^2=u^2+2as

Put the value in the equation

v^2=0+2\time2.25\times525

v=\sqrt{2\times2.25\times525}

v=48.60\ m/s

We need to calculate the maximum height with initial velocity

Using equation of motion

v^2=u^2-2gh

h=\dfrac{v^2-u^2}{-2g}

Put the value in the equation

h=\dfrac{0-(48.60)^2}{-2\times9.8}

h=120.50\ m

The total height  reached by the rocket is

h'=s+h

h'=525+120.50

h'=645.5\ m

Hence, The maximum height that the rocket reaches is 645.5 m.

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Light rays bend as they pass from air into water at an angle (not 90 degrees). this is refraction. air water normal incident ray
xxTIMURxx [149]

When light ray pass from air into water, its speed and wavelength change only the frequency of the light doesn't change.

Light travels slower in a medium of higher refractive index. It bends because of this change in speed. The wavelength of light also changes in order to maintain the constant frequency.

7 0
3 years ago
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
3 years ago
A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly
Elden [556K]

Answer:

a. B= 9.45 \times10^{-3} T

b. B= 0.820 T

c. B= 0.0584 T

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

J= \frac {-I_{c}}{\pi(c^{2}-b^{2}  ) }}

J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}

B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\

B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0
3 years ago
Because of your success in physics class you are selected for an internship at a prestigious bicycle company in its research and
tiny-mole [99]

To develop the problem it is necessary to apply the equations related to the moment of inertia.

The given values can be defined as,

M = 1.0 kg

r = 0.5 m

m = 10 g

I = 0.280 kg.m^2

According to the definition of the moment of inertia applied to the exercise we can arrive at the equation that,

I = I_{rim} + n * I_{spoke}

Where n is the number of spokes necessary to construct the wheel.

I_{rim} = M*r^2 = 1.0 * 0.5^2

I_{spoke} = \frac{1}{3} * m * r^2 = \frac{1}{3}* 10 * 10^-3 * 0.5^2

Replacing the values at the general equation we have,

0.280 = 1.0 * 0.5^2 + n * (1/3 * 10 * 10^-3 * 0.5^2 )

Solving for n,

n = 36

Therefore the number of spokes necessary to construct the wheel is 36

PART B) The mass of the wheel is given by the sum of all masses and the total spokes, then

M_w= M + n*m

M_w = 1.0 + 36* 10 * 10^{-3} Kg

M_w = 1.36 Kg

Therefore the mass of the wheel must be of 1.36Kg

4 0
4 years ago
Help Please I Don't Know.
Tom [10]
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3CO2: carbon, oxygen
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2)
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8 0
3 years ago
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