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ivann1987 [24]
3 years ago
7

Space-shuttle astronauts experience accelerations of about 35 m/s2 during takeoff. What force does a 75 kg astronaut experience

during an acceleration of this magnitude?
Physics
1 answer:
amm18123 years ago
5 0

Answer:

<h3>The answer is 2625 N</h3>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 75 × 35

We have the final answer as

<h3>2625 N</h3>

Hope this helps you

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In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact
OLga [1]

Answer:

v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}

so we can now solve this for k, which is the constant of the spring.

k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

which yields:

V_{B}=3.73m/s

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

\sum F=ma

so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

a=\frac{F_{s}+f-W}{m}

or

a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

which yields:

3.35m/s^{2}

8 0
4 years ago
State Pascal''s principle. Give and example of its use
BabaBlast [244]
Well, it is also known as the transmission of fluid pressure. So, it is a principle in fluid mechanics that says that pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such as the the pressure variations.
4 0
4 years ago
(V) What are the objectives of the space mission?​
Svetach [21]

Explanation:

I hope this helps, I got this from another person

3 0
3 years ago
A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there
____ [38]

<em>Answer</em>


0.6 teslas


<em>Explanation</em>

When a conductor is inside a magnetic field it experiences a force given by;

Force = ILBsinθ


Where I⇒ current

L ⇒length of the conductor

B ⇒ magnetic field strength

θ ⇒ Angle between the conductor and magnetic field.


F = ILBsinθ


When θ = 90°, Then sin 90 =1 and the formula becomes;

F =ILB

3 = 10 × 0.5 × B

3 = 5B

B = 3/5

= 0.6


magnetic field strength = 0.6 teslas


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3 years ago
Where were the girls heading when the car broke down?<br> in the movie *hidden figures*
Ilia_Sergeevich [38]

Explanation:

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3 years ago
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