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3 years ago

A cat is running at 24 m/s. It then accelerates at 7 m/s2. How long will it take the cat to reach a speed of 49 m/s?

Physics

1 answer:

kozerog [31]3 years ago

8 0

Answer:

t should be 3.57 second

Explanation:

Formula used is v = u+at

In which v is final velocity, u is initial velocity, a is acceleration and t is time.

Substitute each of the info given into the formula and calculate.

49 = 24 + (7)t

t = 3.57s

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The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m

butalik [34]

$\sf\underline{Solution:}$

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

$\sf{Now,according \: to \:the\:question:}$

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

☆ Substitute the required values in the given formula-

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0

2 years ago

A 2.10 cm × 2.10 cm square loop of wire with resistance 1.30×10−2 Ω has one edge parallel to a long straight wire. The near edge

Troyanec [42]

Answer:

$I_{l} =44.84 \mu A$

Explanation:

given,

side of square loop = a = 2.10 cm

Resistance of the wire = 1.30×10⁻² Ω

Length of the loop = c = 1.10 cm

rate of increasing current = 130 A/s

$\phi = \dfrac{\mu_0Ib}{2\pi}(ln(\dfrac{c+a}{c}))$

$\dfrac{d\phi}{dt}= \dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{V}{R}$

$I_{l} = \dfrac{1}{R}\dfrac{d\phi}{dt}$

$I_{l} = \dfrac{1}{R}\dfrac{\mu_0b}{2\pi}\dfrac{dI}{dt}(ln(\dfrac{c+a}{c}))$

$I_{l} = \dfrac{1}{1.3 \times 10^{-2}}\dfrac{4\pi\times 10^{-7}\times 0.021}{2\pi}\times 130\times (ln(\dfrac{0.011+0.021}{0.011}))$

$I_{l} =44.84 \times 10^{-6}\A$

$I_{l} =44.84 \mu A$

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3 years ago

Physical Science convert 5.3 hl to dal

Nikitich [7]

Answer: 53 dal

Explanation: 5.3x10=53, to get dal you have to multiply by 10 to get 5.3x10=53.

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2 years ago

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hichkok12 [17]

Answer:

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if mass is 3, and velocity is 3 (3^2 = 9)

then KE increases by a factor of 27

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goldenfox [79]

Answer:

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Explanation:

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