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wlad13 [49]
2 years ago
8

Two identical stunt professionals A and B stand on the roof of building 1. Person A steps off of the roof and falls vertically o

nto the solid ground. At the same instant, person B steps off the roof while holding onto a rope with its opposite end is fixed to building 2 of the same height. The rope is initially horizontal, and after swinging downward almost the same vertical distance as A fell, person B collides with the solid wall of building 2.
A. Which person will sustain more serious injuries, or will both sustain injuries of the same seriousness? Explain your reasoning.
Physics
1 answer:
Vladimir [108]2 years ago
6 0

Answer:

Both of the stunt professionals will sustain injuries of the same seriousness

Explanation:

We are being told that both stunt professionals are standing from the same height, therefore they will attain the same  equivalent speed at the bottom if we are to look at it from the principle of conservation of energy.

Now; According to principle of momentum; the momentum at which the first stunt professional A hits the ground be equal as the momentum with which stunt professional B will hit the wall.

Thus; both of the stunt professionals will sustain injuries of the same seriousness

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An electron moving at right angles to a 0.1 T magnetic field experiences an acceleration of 6 × 1015 m.s-2. What is the speed of
GaryK [48]

Explanation:

It is given that,

Magnetic field, B = 0.1 T

Acceleration, a=6\times 10^{15}\ m/s^2

Charge on electron, q=1.6\times 10^{-19}\ C    

Mass of electron, m=9.1\times 10^{-31}\ kg    

(a) The force acting on the electron when it is accelerated is, F = ma

The force acting on the electron when it is in magnetic field, F=qvB\ sin\theta

Here, \theta=90

So, ma=qvB

Where

v is the velocity of the electron

B is the magnetic field

v=\dfrac{ma}{qB}

v=\dfrac{9.1\times 10^{-31}\ kg\times 6\times 10^{15}\ m/s^2}{1.6\times 10^{-19}\ C\times 0.1\ T}

v = 341250  m/s

or

v=3.41\times 10^5\ m/s

So, the speed of the electron is 3.41\times 10^5\ m/s

(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.

7 0
2 years ago
) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force
Nana76 [90]
Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others. 
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
F_x=F_3+F_2cos(45^{\circ})
F_y=F_1+F_2sin(45^{\circ})
Let's figure out each component:
F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}
Total force acting on the charge would be:
F=\sqrt{F_x^2+F_y^2}
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N
F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N
Now we can find the total force acting on a single charge:
F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N
As said before, intensity of the force acting on charges is the same for all of them.

5 0
2 years ago
A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
LUCKY_DIMON [66]

Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Put the value into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

v=ki

v=\dfrac{kh}{l}

Put the value into the formula

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test is 0.0187 cm/s.

(b). We need to calculate the volume of solid in the ample

Using formula of volume

V_{s}=\dfrac{M_{s}}{V_{s}}

Put the value into the formula

V_{s}=\dfrac{4950\times10^{-3}}{2710}

V_{s}=1826.56\ cm^3

We need to calculate the volume of the soil specimen

Using formula of volume

V=A\times L

Put the value into the formula

V=\dfrac{\pi(17.5)^2}{4}\times17.5

V=4209.24\ cm^3

We need to calculate the volume of the voids

V_{v}=V-V_{s}

Put the value into the formula

V_{v}=4209.24-1826.56

V_{v}=2382.68\ cm^3

We need to calculate the seepage velocity

Using formula of velocity

Av=A_{v}v_{s}

v_{s}=\dfrac{Av}{A_{v}}

v_{s}=\dfrac{V}{V_{v}}\times v

Put the value into the formula

v_{s}=\dfrac{4209.24}{2382.68}\times0.0187

v_{s}=0.0330\ cm/s

The seepage velocity is 0.0330 cm/s.

Hence, (a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

8 0
2 years ago
A 15.0 g bullet traveling horizontally at 865 m>s passes through a tank containing 13.5 kg of water and emerges with a speed
Murrr4er [49]

Answer:

The rise in temperature is 0.06 K.

Explanation:

mass of bullet, m = 15 g

initial speed, u = 865 m/s

final speed, v = 534 m/s

mass of water, M = 13.5 kg

specific heat of water, c = 4200 J/kg K

The change in kinetic energy

K = 0.5 m(u^2 - v^2)\\\\K = 0.5\times 0.015\times (865^2-534^2)\\\\K = 3473 J

According to the conservation of energy, the change in kinetic energy is used to heat the water.

K = m c T

where, T is the rise in temperature.

3473 = 13.5 x 4200 x T

T = 0.06 K

3 0
1 year ago
In batting practice a batter hits a ball that is sitting at rest on top of a post. The ball leaves the post with a horizontal sp
Effectus [21]

Answer:

65.3

Explanation:

W = (1/2)mv^2

W = (1/2)(0.145kg)(30m/s)^2

5 0
2 years ago
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