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2 years ago

10

Myron was trying to light a match. The first time he struck the match, the match did not light. The second time he tried, the ma

tch lit successfully.
Which most likely occurred to make the match light?

Myron increased the friction by decreasing the surface area.
Myron increased the friction by increasing speed.
Myron decreased the friction by making the match aerodynamic.
Myron decreased the friction by lubricating the match.

2 answers:

Andru [333]2 years ago

8 0

Answer:

B. Myron increased the friction by increasing speed.

Explanation:

i got it correct on edge

Volgvan2 years ago

4 0

Answer:

The answer is B i think

Explanation:

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A golfer hits a golf ball with a club head velocity of 94 mph. Mass of golf club head (m): 190 g Mass of golf ball (ms): 46 g Co

MAXImum [283]

Answer:

Velocity of golf ball≅ 12.87mph

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= $\frac{Vf2- Vf1}{Vi1 - Vi2}$

0.83= $\frac{Vf2 - Vf1}{94 - 0}$

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph

7 0

3 years ago

Vesna Vulovic survived the longest fall on record without a parachute when her plane exploded and she fell 5 miles, 733 yards. W

Valentin [98]

Answer:

8717 meters.

Explanation:

We need to know the conversion factors. We know that:

1 mile = 1609.34 meters

1 yard = 0.9144 meters

This means that:

$\frac{1609.34 meters}{1 mile}=1$

$\frac{0.9144 meters}{1 yard}=1$

It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.

We will convert first the miles, then the yards, and add them up.

$5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters$

$733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters$

So total distance is the sum of these, 8717 meters.

8 0

3 years ago

4 The temperature at the center of the sun is 1.55 x 10' K. Express this in degrees

Sloan [31]

Answer:

The answer is D

Explanation:

3 0

3 years ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

3 years ago

Visually, how can you distinguish between an ac and dc power supply?

Temka [501]

Answer:

By the use of slow motion camera.

Explanation:

Visually, it is very hard to differentiate between an ac and dc power supply. But Since, we that In Ac supply polarity changes 100 times in a second ( because frequency of ac supply is 50 Hz generally). Whereas, Dc gives a steady power supply. So, in slow motion camera we can easily capture the flickering light tubes which won't happen in case of dc supply.

5 0

3 years ago

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