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2 years ago

50 POINTS!!!!!!!! In a GUT (Grand Unified Theory), what is "unified"?

Physics

1 answer:

melomori [17]2 years ago

7 0

Subatomic particles

Explanation:

Grand unification theory or GUT is a model that tries to describe the universe.

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15. A volleyball player who weighs 650 Newtons jumps 0.500 meters vertically off the floor. What is her kinetic energy just befo

allochka39001 [22]

<h2>Kinetic energy just before hitting the floor is 324.57 J</h2>

Explanation:

Weight of volleyball player = 650 N

That is

Mass x Acceleration due to gravity = 650

Mass x 9.81 = 650

Mass = 66.26 kg

We also have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Final velocity, v = ?

Displacement, s = 0.5 m

Substituting

v² = u² + 2as

v² = 0² + 2 x 9.81 x 0.5

v = 3.13 m/s

Velocity with which he lands on ground is 3.13 m/s

We have kinetic energy = 0.5 x Mass x Velocity²

Substituting

Kinetic energy = 0.5 x 66.26 x 3.13²

Kinetic energy = 324.57 J

Kinetic energy just before hitting the floor is 324.57 J

3 0

3 years ago

It is now virtually certain that the dominant cause of recent ice ages is

Anarel [89]

<span>Variations in Earth-Sun orbital relationships.</span>

5 0

3 years ago

A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

Lemur [1.5K]

Answer:

$V_2 = 0.125 m^3$

Work done = = 5 kJ

Explanation:

Given data:

volume of nitrogen $v_1 = 0.08 m^3$

$P_1 = 150 kPa$

$T_1 = 200 degree celcius = 473 Kelvin$

$P_2 = 80 kPa$

Polytropic exponent n = 1.4

$\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}$

putting all value

$\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}$

$\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS$

polytropic process is given as

$P_1 V_1^n = P_2 V_2^n$

$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

$V_2 = 0.125 m^3$

work done $= \frac{P_1 V_1 -P_2 V_2}{n-1}$

$= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}$

= 5 kJ

4 0

3 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s.

marishachu [46]

Answer:

a) the magnitude of the force is

F= Q( $\frac{kqs}{r^3}$ ) and where k = 1/4πε₀

F = Qqs/4πε₀r³

b) the magnitude of the torque on the dipole

τ = Qqs/4πε₀r²

Explanation:

from coulomb's law

E = $\frac{kq}{r^{2} }$

where k = 1/4πε₀

the expression of the electric field due to dipole at a distance r is

E(r) = $\frac{kp}{r^{3} }$ , where p = q × s

E(r) = $\frac{kqs}{r^{3} }$ where r>>s

a) find the magnitude of force due to the dipole

F=QE

F= Q( $\frac{kqs}{r^3}$ )

where k = 1/4πε₀

F = Qqs/4πε₀r³

b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces

τ = F sinθ × s

θ = 90°

note: sin90° = 1

τ = F × r

recall F = Qqs/4πε₀r³

∴ τ = (Qqs/4πε₀r³) × r

τ = Qqs/4πε₀r²

8 0

3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago