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sattari [20]
3 years ago
13

A rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of

3.25 metric tons to a speed of 10,000 m/s. It has an engine and fuel designed to produce an exhaust speed of 2400 m/s. How much fuel plus oxidizer is required (in metric tons)
Physics
1 answer:
choli [55]3 years ago
4 0

Answer:

<em>13.54 tons</em>

Explanation:

Let f be the amount of fuel oxidizer needed

v be the speed

The relationship between them is inverse in nature i.e

f ∝ 1/v

f = k/v

If a rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of 3.25 metric tons to a speed of 10,000 m/s, then f = 3.25 when v  = 10,000

Substitute and get k

k = fv

k = 3.25 * 10,000

k = 32500

To get the amount of fuel oxidizer required to produce a speed of 2400m/s, we will find f when v = 2400m/s

Recall that f = k/v

f = 32500/2400

f = 13.54 metric tons

<em>Hence the fuel plus oxidizer that will be required is 13.54 tons</em>

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larisa [96]

Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

x = 83.23 m

Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

6 0
3 years ago
What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?
Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u =  6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

4 0
3 years ago
Tapping the surface of a pan of water generates 17.5 waves per second. If the wavelength of each wave is 45 cm, what is the spee
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Answer:

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Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

5 0
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trasher [3.6K]

Answer:

Belgium

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These are the ones that are in the High Productivity chart, but not in the HDI chart

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That is false

I hope this helps!
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