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2 years ago

Help please! Will give brainliest to first correct answer!

Physics

1 answer:

Olegator [25]2 years ago

8 0

Answer:

1) Addition of a catalyst

2) To change the reaction rate of slope B to look like slope A, simply add a catalyst to speed up the rate of reaction, giving you a higher amount of products in a shorter amount of time (line A)

Explanation:

1 and 2)Two things can alter the rate of a reaction, either the addition of a catylist which will not alter the composition of the products or reactants, but will accelerate the reaction time, or an increase in temperature will also increase the rate at which a reaction will occur.

You could choose temperature also and have the same result, it's your choice both are correct, but catalyst is the easiest.

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A transformer consists of 290 primary windings and 824 secondary windings. Part A: If the potential difference across the primar

jasenka [17]

Answer:

Part 1) Voltage in secondary windings is 61.08 Volts

Part 2) Current in secondary windings is 0.53 Amperes

Explanation:

The potential developed in the primary and secondary winding of a transformer are related as

$\frac{N_{p}}{N_{s}}=\frac{V_{p}}{V_{s}}$

where

Np no of turns in primary coil

Ns no of turns in secondary coil

Vp Voltage of turns in primary coil

Vs Voltage of turns in secondary coil

Applying values in the formula we get

$\frac{290}{824}=\frac{21.5}{V_{s}}\\\\\therefore V_{s}=21.5\times \frac{824}{290}=61.08V$

Part 2)

Using Ohm's law the current is given by

$I=\frac{V_{s}}{R}\\\\I=\frac{61.089}{115}=0.53A$

5 0

2 years ago

answer this question: When you can not see what is taking place, but other senses indicate occurrences. This is called _________

Phoenix [80]

This is called an indirect observation.

5 0

3 years ago

A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 34.5 kg.

rodikova [14]

Answer:

the moment of inertia of the merry go round is 38.04 kg.m²

Explanation:

We are given;

Initial angular velocity; ω_1 = 37 rpm

Final angular velocity; ω_2 = 19 rpm

mass of child; m = 15.5 kg

distance from the centre; r = 1.55 m

Now, let the moment of inertia of the merry go round be I.

Using the principle of conservation of angular momentum, we have;

I_1 = I_2

Thus,

Iω_1 = I'ω_2

where I' is the moment of inertia of the merry go round and child which is given as I' = mr²

Thus,

I x 37 = ( I + mr²)19

37I = ( I + (15.5 x 1.55²))19

37I = 19I + 684.7125

37I - 19 I = 684.7125

18I = 684.7125

I = 684.7125/18

I = 38.04 kg.m²

Thus, the moment of inertia of the merry go round is 38.04 kg.m²

7 0

3 years ago

Read 2 more answers

Do cheese and chips good?

olga_2 [115]

Answer:

YES VERY GOOD

Explanation:

6 0

2 years ago

Read 2 more answers

when earth and moon are separated by a distance of 3.84×10^8 meters, the magnitude of the gravitational force of attraction betw

sammy [17]

The gravitational force would be $8\cdot 10^{20} N$

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this case, we can write the gravitational force between the Moon and the Earth as:

$F=G\frac{M m}{r^2}$

where M is the Earth's mass and m is the Moon's mass.

We know that:

$r=3.84\cdot 10^8 m$ is their separation

$F=2.0 \cdot 10^{20}N$ is the force between them

So we can find:

$GMm = Fr^2 = (2.0\cdot 10^{20})(3.84\cdot 10^8)^2=2.95\cdot 10^{37} Nm^2$

Now, we are asked to find the gravitational force when the Earth and the Moon are separated by

$r'=1.92\cdot 10^8 m$

In this case the new force would be

$F'=G\frac{M m}{r'^2}$

And substituting the values of (GMm) found previously and r', we find

$F'=\frac{2.95\cdot 10^{37}}{(1.92\cdot 10^8)^2}=8\cdot 10^{20} N$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

6 0

3 years ago