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3 years ago

What inspired Ronald McNair to do science

Physics

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

Answer:

While working as a staff physicist at hughes Research Laboratories McNair learned that the National Aeronautics and Space Administration (NASA) was looking for scientist to join the shuttle program;)

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0

2 years ago

When an object of weight w is suspended from the center of a massless string?

igomit [66]

I attached a picture of the diagram associated with this question.

Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ

The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ

4 0

3 years ago

Read 2 more answers

‼️‼️ Please help, urgent ‼️‼️ (check photo)

Alex787 [66]

Answer: The force constant k is 10600 kg/s^2

Step by step:

Use the law of energy conservation. When the elevator hits the spring, it has a certain kinetic and a potential energy. When the elevator reaches the point of still stand the kinetic and potential energies have been transformed to work performed by the elevator in the form of friction (brake clamp) and loading the spring.

Let us define the vertical height axis as having two points: h=2m at the point of elevator hitting the spring, and h=0m at the point of stopping.

The total energy at the point h=2m is:

$E_{tot}=E_{kin}+E_{pot}\\E_{tot}= \frac{1}{2}mv^2+mg\Delta h = \frac{1}{2}2000 kg 4^2\frac{m^2}{s^2}+2000kg\, 9.8\frac{m}{s^2}2m=55200\,kg\frac{m^2}{s^2}$