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2 years ago

You run away from a plane mirror at 2.30 m/s. At what speed does your image move away from you?

Physics

1 answer:

ivanzaharov [21]2 years ago

5 0

Answer:

4.60m/s

.............

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Find the velociity of a car which travels 35 m to the right over a period of 40 seconds

GalinKa [24]

Answer:

the velocity of the car is 0.875 m/s

Explanation:

$v = d \div t \\ v = 35m \div 40sec = 0.875 \: m \: per \: sec$

therefore the V of car is 0.875 m

8 0

2 years ago

A 26.2-kg dog is running northward at 3.02 m/s, while a 5.30-kg cat is running eastward at 2.74 m/s. Their 65.1-kg owner has the

REY [17]

Answer:

Angle with the +x axis is θ = 79.599degree

Then the velocity of owner = 1.235m/s

Explanation:

Given that the mass of dog is m1 =26.2 kg

velocity of dog is u1 = 3.02 m/s (north)

mass of cat is m2 = 5.3 kg

velocity is u2 = 2.74 m/s (east )

Mass of owner is M = 65.1 kg

Consider the east direction along +x axis andnorth along +y

momentum of dog is Py = m1 x u1

= 79.124 kg.m/s (j)

momentum of cat is Px = m2 x u2

= 14.522 kg.m/s (i)

Then the net magnitude of momentum is P = (Px2 + Py2)1/2

= 80.445

Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree

Then the velocity of owner is v = P / M = 1.235 m/s

3 0

2 years ago

It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, andthat the value of n in the expon

trapecia [35]

Answer: $8.76\times 10^{-3} min^{-1}$

Explanation:

Given

n=5

0.3 fraction recrystallize after 100 min

According to Avrami equation

$y=1-e^{-kt^n}$

where y=fraction Transformed

k=constant

t=time

$0.3=1-e^{-k(100)^5}$

$e^{-k(100)^5} =0.7$

Taking log both sides

$-k\cdot (10^{10}=\ln 0.7$

$k=3.566\times 10^{-11}$

At this Point we want to compute $t_{0.5}\ i.e.\ y=0.5$

$0.5=1-e^{-kt^n}$

$0.5=e^{-kt^n}$

$0.5=e^{-3.566\times 10^{-11}\cdot (t)^5}$

taking log both sides

$\ln 0.5=-3.566\times 10^{-11}\cdot (t)^5$

$t^5=1.943\times 10^{10}$

$t=114.2 min$

Rate of Re crystallization at this temperature

$t^{-1}=8.76\times 10^{-3} min^{-1}$

3 0

2 years ago

How can an animal regeneration produce two results?

GenaCL600 [577]

In biology, regeneration refers to the process by which plants and animals replace lost or damaged parts by growing them anew. Some animals can regenerate their limbs, tails, or even parts of internal organs, such as the liver. In plant regeneration, neighboring cells replace missing tissue.

8 0

2 years ago

Read 2 more answers

Which of the following is an example of a chemical change?

ss7ja [257]

Answer:

Burning Paper

Explanation:

This is a chemical reaction, because new substances are formed

6 0

2 years ago