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2 years ago

6

أسقط عامل حجرة من سطح بناية سقوطأ حرة. أوجد ما يلي :

1 answer:

Ilia_Sergeevich [38]2 years ago

7 0

Answer:

??? i don't no what you just said

Explanation:

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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

3 0

3 years ago

Which is a product of nuclear fusion?

Tju [1.3M]

Answer:

Creation of energy by joining the nuclei of two hydrogen atoms to form helium

7 0

3 years ago

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An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-

kirill [66]

Answer:0.478 c

Explanation:

Given

mass of lighter Particle $(m_1)=3\times 10^{-28} kg$

mass of heavier Particle $(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle $(v_1)=0.834 c$

Let speed of heavier particle $=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

$8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

$v_2=0.478 c$

3 0

3 years ago

Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated brakes. For a simpl

o-na [289]

Answer:

404.4 m

Explanation:

Converting the initial speed from km/h to m/s then

$140\times \frac {1000m}{3600s}=38.88888889 m/s \approx 38.89 m/s$

The acceleration is resolved as shown in the figure hence

deceleration of the truck along the inclined plane will be

$a=-g sin \theta$ where g is acceleration due to gravity

Substituting g with $9.81 m/s^{2}$ then

$a=-9.81 m/s^{2} sin 11^{\circ}=-1.871836245\approx -1.87 m/s^{2}$

Using kinematic equation

$v^{2}=u^{2}+2as$ and making s the subject then

$s=\frac {v^{2}-u^{2}}{2a}$ where v and u are final and initial velocities respectively

Substituting 0 for v, 38.89 m/s for u and $-1.87 m/s^{2}$ then

$s=\frac {0^{2}-38.89^{2}}{2\times -1.87}=404.3936096 m\approx 404.4 m$

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3 years ago

At which of the following angles will the sunlight received at a location on Earth spread out over the largest area?

yarga [219]

10 degrees, because 0 degrees will receive the most sunlight, and 10 degrees is closest to 0 degrees.

5 0

3 years ago

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