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tia_tia [17]
3 years ago
11

Airplane flight recorders must be able to survive catastrophic crashes. Therefore, they are typically encased in crash-resistant

steel or titanium boxes that are subjected to rigorous testing. One of the tests is an impact shock test, in which the box must survive being thrown at high speeds against a barrier. A 52-kg box is thrown at a speed of 400 m/s and is brought to a halt in a collision that lasts for a time of 6.5 ms. What is the magnitude of the average net force that acts on the box during the collision
Physics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

3.2 MN

Explanation:

Given that

Mass of the box, m = 52 kg

Initial velocity of the box, u = 400 m/s

Final velocity of the box, v = 0 m/s

Time taken for the collision, t = 0.0065 s

Using the equation of motion

V = u + at, we turn around and make acceleration, a the subject of the formula. Now we have,

a = (v - u) / t

a = (0 - 400) / 0.0065

a = 61538.5 m/s²

The acceleration(which is negative acceleration or retar.dation actually) is 61538.5 m/s². We then proceed to is this acceleration in the basic Force equation, to get the magnitude of force needed.

Remember,

Force = mass * acceleration

F = ma, we already have our mass and acceleration, all we do is multiply

F = 52 * 61538.5

F = 3200002 N or 3.2 MN

Therefore, the magnitude of the force that acts on the box during collision is 3.2 MN

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What is the impulse of a 3 kg object that starts from rest and moves to 20 m/s?
IrinaK [193]

Answer:

The impulse on the object is 60Ns.

Explanation:

Impulse is defined as the product of the force applied on an object and the time at which it acts. It is also the change in the momentum of a body.

F = m a

F = m(\frac{v_{2}  - v_{1} }{t})

⇒     Ft = m(v_{2} - v_{1})

where: F is the dorce on the object, t is the time at which it acts, m is the mass of the object, v_{1} is its initialvelocity and v_{2} is the final velocity of the object.

Therefore,

impulse = Ft = m(v_{2} - v_{1})

From the question, m = 3kg, v_{1} = 0m/s and v_{2} = 20m/s.

So that,

Impulse = 3 (20 - 0)

             = 3(20)

             = 60Ns

The impulse on the object is 60Ns.

8 0
2 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
3 years ago
Read 2 more answers
An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s
max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

F_e=qE=ma             (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

x_{max}=v_o\sqrt{\frac{2d}{a}}             (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm

The horizontal distance traveled by the electron is 9.5cm

4 0
3 years ago
3. What is the frequency of a wave that has a wave speed of 20 m/s and a wavelength of 0.50 m?
bonufazy [111]

Explanation:

everything can be found in the picture

3 0
3 years ago
Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of
Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0
3 years ago
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