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3 years ago

WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!

Physics

1 answer:

Nesterboy [21]3 years ago

3 0

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

(comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

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Jake is in chemistry class. He makes a list of the chemicals his instructor described and the properties of each.

koban [17]

Silver: bonds with other atoms because of the weak forces of the valence electrons
FALSE - The strong forces of the valence electrons is actually the reason why silver bonds with other atoms.

Water: bonds allow for liquid state at room temperature and prevent conduction
FALSE - Water is a good conductor.

Carbon: bonds with other atoms through strong shared electrical bonds
TRUE - Carbon shares covalent bonds with other atoms.

Niobium: bonds allow for a strong conductivity found in stainless steel
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3 years ago

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2. A powerful experimental sewing machine is powered by a mass-spring system. This

Alexus [3.1K]

We have that the Number of stitches per sec and he mass of oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

Generally the equation for the Time t is mathematically given as

$T=2\pi\sqrt{\frac{m}{k}}$

Therefore

$0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}$

m=2.033e-5kg

For more information on Mass visit

brainly.com/question/15959704

7 0

2 years ago

In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)

salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

$m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm$

Object distance is 9 cm

$v=-4\times 9=-36\ cm$

Image distance is -36 cm (other side of object)

7 0

3 years ago

Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

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3 years ago

Timmy drove 2/5 of a journey at an average speed of 20 mph.

mixer [17]

Answer:

4hr

Explanation:

5 0

3 years ago

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