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2 years ago

The key difference between the binomial and hypergeometric distribution is that

Physics

1 answer:

Ivenika [448]2 years ago

5 0

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

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A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring

AleksandrR [38]

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

$v^2-u^2=2as$

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

$0-(17.7)^2=2a\times 33.1$

$a=\dfrac{-(17.7)^2}{33.1}$

$a=-9.46\ m/s^2$

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

$F = ma$

$F =1400\times(-9.46)$

$F=-13244\ N$

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

$|F| =13244\ N$

Hence, The magnitude of the horizontal net force is 13244 N.

5 0

3 years ago

If a ball has a velocity of 5 m/s and a kinetic energy of 1000 J, what is its mass?

zheka24 [161]

Answer: I don't know how to do this

Explanation: sorry I am not sure.

7 0

2 years ago

A force of 100 N is applied to a crate at an angle of 60° to the horizontal surface. The force causes the crate to move 2 m acro

MAVERICK [17]

Answer: B 100 J

Explanation: Pic Attached. Hope This Helps.

4 0

2 years ago

Be sure to answer all parts. By what factor does the fraction of collisions with energy equal to or greater than activation ener

qwelly [4]

Answer:

Factor = 8.77

Explanation:

Fraction of collision with energy greater than or equal to the activation energy can be given by the formula:

$f = \exp(\frac{-E_{a} }{RT} )$

$E_{a} = Activation Energy\\E_{a} = 100 kJ /mol\\E_{a} = 10^{5} J /mol\\$

R = 8.314 J/mol.K

When Temperature = 34⁰C = 34 + 273 = 307 K

$f_{1} = \exp(\frac{-10^{5} }{8.314 * 307} )\\f_{1} = \exp(\frac{-10^{5} }{2552.398} )\\f_{1} = \exp(-39.18)\\f_{1} = 964.59 * 10^{-20}$

When Temperature = 52⁰C = 52 + 273 = 325 K

$f_{2} = \exp(\frac{-10^{5} }{8.314 * 325} )\\f_{2}= \exp(\frac{-10^{5} }{2702.05} )\\f_{2}= \exp(-37.009)\\f_{2} = 8456.6 * 10^{-20}$

$\frac{f_{2}}{f_{1}}= \frac{8456.6 * 10^{-20} }{964.59 * 10^{-20} }$

Factor = 8.77

8 0

2 years ago

An object's true weight is 123 N. When it is completely submerged in water, its

SashulF [63]

Object true weight is given as

$mg = 123 N$

now we know that g = 9.8 m/s^2

$m* 9.8 = 123$

$m = \frac{123}{9.8} = 12.55 kg$

now when it is complete submerged in water its apparent weight is given as 82 N

apparent weight = weight - buoyancy force

apparent weight = 82 N

weight = 123 N

now we have

82 = 123 - buoyancy force

buoyancy force = 123 - 82 = 41 N

now we also know that buoyancy force is given as

$F_b = p_{liq}Vg$

$41 = 1000*V*9.8$

$V = \frac{41}{1000*9.8}$

$V = 4.18 * 10^{-3} m^3$

now as we know that mass of the object is 12.55 kg

its volume is 4.18 * 10^-3 m^3

now we know that density will be given as mass per unit volume

$density = \frac{m}{V}$

$density = \frac{12.55}{4.18*10^{-3}$

$density = 3002.4 kg/m^3$

so here density of object is 3002.4 kg/m^3

7 0

2 years ago

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