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3 years ago

If a cars speed increases from 18 m/s to 26 m/s in 25 s, her acceleration is.

Physics

1 answer:

Step2247 [10]3 years ago

8 0

Answer:

0.32

Explanation:

u=18m/s

v=26m/s

t=25s

a=v-u/t

a=26-18/25

a=0.32

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3 years ago

PLEASE help with this question!

Ainat [17]

Answer:

the answer might the number 2

Explanation:

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3 years ago

Was the first chemist to organize elements by atomic number

Arturiano [62]

Answer:

Dimitri Mendeleev

Explanation:

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3 years ago

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A 50.0-kg child stands at the rim of a merry-go-round of radius 2.00 m, rotating in such a way that it makes one revolution in 2

Vaselesa [24]

Answer:

2.01

Explanation:

First, we need to find the centripetal acceleration.

We're given that the merry go round rotates 1 revolution in 2.09 seconds. Converting to rpm, we know that it rotates 30 revolution per minute

Now this speed gotten in rpm will be converted to m/s, to ease the calculation

30 rpm = πdN/60 m/s

30 rpm = (3.142 * 4 * 30)/60

30 rpm = 377.04/60

30 rpm = 6.284 m/s

a(c) = v²/r

a(c) = 6.284²/2

a(c) = 39.49 / 2

a(c) = 19.74 m/s²

F = ma

F = 50 * 19.74

F = 987 N

Also, Normal Force, F(n) =

F(n) = mg

F(n) = 50 * 9.81

F(n) = 490.5

We then use this to find the coefficient of static friction, μ

μ = F/F(n)

μ = 987 / 490.5

μ = 2.01

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3 years ago

The rate at which a metal alloy oxidizes in an oxygen-containing atmosphere is a typical example of the practical utility of the

Ray Of Light [21]

Answer:

The activation energy is $Q = 328.31 \ K J/mol$

Explanation:

From the question we are told that

The rate constant is k

at the temperature $T_1 = 300 = 300 + 273 = 573 \ K$

The value of k is $k_1 = 1.05 *10^{-8} \ kg /m^4 \cdot s$

at temperature $T_2 = 400 ^oC = 400 + 273 = 673 \ K$

The value of k is $k_2 = 2.95 *10^{-4} \ kg /m^4 \cdot s$

The rate constant is mathematically represented as

$k = Ce^{- \frac{Q}{RT} }$

Where Q is the activation energy

R is the ideal gas constant with a value of $R = 8.314 \ J /mol \cdot K$

C is a constant

T is the temperature

For the first rate constant

$k_1 = Ce ^{-\frac{Q}{RT_1} }$

For the second rate constant

$k_2 = Ce ^{-\frac{Q}{RT_2} }$

Now the ratio between the two given rate constant is

$\frac{k_1 }{k_2} = e^{(\frac{Q}{R} [\frac{1}{\frac{T_2 - 1}{T_1} } ] )}$

=> $ln [\frac{k_1}{k_2} ] = \frac{Q}{R} * [\frac{1}{\frac{T_2 -1}{T_1} } ]$

substituting values

$ln [\frac{1.05 *10^{-8}}{2.95 *10^{-4}} ] = \frac{Q}{8.314} * [\frac{1}{\frac{673 -1}{573} } ]$

=> $Q = 328.31 \ K J/mol$

7 0

3 years ago