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3 years ago

Please help need answer asp

Physics

2 answers:

MrMuchimi3 years ago

6 0

Flexibility helps a joint to move in any direction

slega [8]3 years ago

4 0

Flexability because when you are bendable you can bend in any direction and that's what flexablitity is

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Safety devices used in electric circuit

Sergeu [11.5K]

Fuses is the answer!

6 0

3 years ago

If you wanted to know how big a tiger is, which measurements would you use?

valina [46]

Answer:

meter, kilogram

Explanation:

Here we want to know how big the tiger is. This means that we want to measure its size and possibly its mass.

The size is actually a measure of the length of the tiger, and length is measured in meters.

The mass of an object, instead, is a measure of the "amount of matter" in the substance, and it is measured in kilograms.

The other options are wrong because:

- The second is the unit of time

- The candela is the unit of the luminous intensity

- The mole is the unit of the amount of substance, and it is used for gases

- The ampere is the unit of the current

5 0

3 years ago

When is the pressure of the man on the ground more – lying position or standing position? Explain.

Dmitrij [34]

It is without a doubt in standing position!
“I will explain if u ask”
(Please mark brainlest if I helped u a bit)

3 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 525 K and again at 545 K . What is the ratio of f at the h

Kazeer [188]

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

$f=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$f_1$ = rate constant at 525K

$K_2$ = rate constant at 545K

$Ea$ = activation energy for the reaction = 185kJ/mol= 185000J/mol (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 525 K

$T_2$ = final temperature = 545 K

Now put all the given values in this formula, we get

$\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}]$

$\log (\frac{f_2}{f_1})=0.6754$

$(\frac{f_2}{f_1})=4.736$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736

8 0

3 years ago

If the rest energies of a positive sigma particle and a lambda particle are 1189.4 and 1115.7 Mev respectively, what is the diff

Lunna [17]

Answer:

1.31022×10⁻³⁴ kg

Explanation:

Difference in energies = 1189.4-1115.7 MeV = 73.7 MeV

Convert this energy to Joules

1 MeV = 1.6×10⁻¹⁹ J

73.7 MeV = 73.7×1.6×10⁻¹⁹ J = 117.92×10⁻¹⁹ J

c = Speed of light = 3×10⁸ m/s

From Einstein's equation

E = mc²

$m=\frac{E}{c^2}\\\Rightarrow m=\frac{117.92\times 10^{-19}}{(3\times 10^8)^2}\\\Rightarrow m=1.31022\times 10^{-34}\ kg$

∴ Difference in their masses is 1.31022×10⁻³⁴ kg

8 0

3 years ago