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2 years ago

Hypersonic scramjet. On March 27, 2004, the United States

Physics

1 answer:

kati45 [8]2 years ago

4 0

25.487 km did the scramjet travel during its 11 second test.

Given that the scramjet travels at a speed seven times that of sound, speed v is equal to 7 x 331 m/s or 2317 m/s.

(A) We require the scramjet's travel time over a distance of

d = 5000 km (5x10^6 m).

Using the formula

time taken = distance traveled/speed,

we can calculate the following:

(5x10^6)/2317 sec

= (5x10^6)/(23175x60) minute

= 5000000/139020 minutes

= 35.97 seconds

As 1 second equals 1/60 of a minute, Therefore, the travel time on Scramjet from San Francisco to New York will be 35.97 minutes.

(B) The distance covered by the Scramjet in 11 seconds must now be determined. as a result,

using speed, distance, and time relation.

Distance = Speed x Time

= 7 x 331 x 11

= 25487 m

Since 1 metre equals 1/1000 of a kilometer, 25487 m of 25.487 km

Learn more about Distance here:

brainly.com/question/15172156

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Distance from center of platform = 1.31 m

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MOI of disk = $\frac{MR^2}{2}$

Moment of inertia of the person and the dog will be mr^2.

Where m and r are different for both the bodies.

So,

Moment of inertia $(I_y_y )$ of the system with respect to the axis yy.

⇒ $I_y_y=I_d_i_s_k + I_m_a_n+I_d_o_g$

⇒ $I_y_y=\frac{M_d_i_s_k(R_d_i_s_k)^2}{2} +M_m(r_c)^2+M_d_o_g(R_c)^2$

⇒ $I_y_y=\frac{129(1.61)^2}{2} +65.7(1.07)^2+27.2(1.31)^2$

⇒ $I_y_y=289.088\ kg.m^2$

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