A policeman kicks in a door with a force of 4500 N. What force does the door apply to the policeman’s leg?

A particle of mass M is moving in the positive x direction with speed v. It spontaneously decays into 2 photons, with the origin

anygoal [31]

Solution :

Mass of the particle = M

Speed of travel = v

Energy of one photon after the decay which moves in the positive x direction = 233 MeV

Energy of second photon after the decay which moves in the negative x direction = 21 MeV

Therefore, the total energy after the decay is = 233 + 21

= 254 MeV

So by the law of conservation of energy, we have :

Total energy before the decay = total energy after decay

So, the total relativistic energy of the particle before its decay = 254 MeV

7 0

2 years ago

A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. if the modulus of elasticity of the material of

Scilla [17]

Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
E=2*10^5 N/mm^2
Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
σ
=force/area
= 20000 N / 314 mm^2
= 6366.2 N/mm^2
= 6370 N/mm^2 (to 3 significant figures)

(ii) Strain
ε
= ratio of extension / original length
= σ / E
= 6366.2 /(2*10^5)
= 0.03183
= 0.0318 (to three significant figures)

(iii) elongation
= ε * L
= 0.03183*1500 mm
= 47.746 mm
= 47.7 mm (to three significant figures)

5 0

2 years ago

A ball rolls 6.0 meters as its speed changes from 15 meters per second to 10 meters per second. What is the average speed of the

antoniya [11.8K]

Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m

The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²

The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s

The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s

Answer: 12.5 m/s

6 0

2 years ago

¿Alguien me puede ayudar? Problema: Un niño le pide gastada a su papá y éste le contesta que le dará los $120 que tiene en su bo

VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

3 0

2 years ago

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0

3 years ago