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KATRIN_1 [288]
3 years ago
13

A bowling ball with a negative initial velocity slows down as it rolls down the lane toward the pins. Is the bowling ballâs acce

leration positive or negative as it rolls toward the pins?
Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer:

positive

Explanation:

The ball is rolling down with a negative velocity, but the velocity is slowing down. therefore the velocity must increase in order for the ball to slow down.

For example let the ball's initial velocity be -15 m/s. and it  is slowing down to let's say -13 m/s. Well this means that it's velocity has increase by 2 m/s. So, its acceleration is positive.

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What type of reaction feels cold to the touch? A. Endothermic, because energy is being released into the surroundings B. Endothe
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Answer:

D. Exothermic, because energy is being absorbed from the surroundings

Explanation:

This is true about the Exothemic reaction due to the fact that, the reaction occurs outside the body. During this reaction, the energy being absorbed <em>from the surrounding environment will hit the body surface thereby creating the coldness due to the heat given out from the body  being minimal.</em>

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Sound waves are mechanical waves in which the particles in the medium vibrate in a direction parallel to the direction of energy
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C longitudnal waves

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Label and describe what is happening in this picture
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3 years ago
A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω. The material constant, β, for this therm
Karo-lina-s [1.5K]

Answer:

the thermistor temperature = 325.68 \ ^0 \ C

Explanation:

Given that:

A thermistor is placed in a 100 °C environment and its resistance measured as 20,000 Ω.

i.e Temperature

T_1 = 100^0C\\T_1 = (100+273)K\\\\T_1 = 373\ K

Resistance of the thermistor R_1 = 20,000 ohms

Material constant \beta = 3650

Resistance of the thermistor R_2 = 500 ohms

Using the equation :

R_1 = R_2  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{R_1}{ R_2} =   \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

Taking log of both sides

In \ \frac{R_1}{ R_2} = In \  \ e^{\beta} (\frac{1}{T_1}- \frac{1}{T_2})

In \ \frac{R_1}{ R_2} = {\beta} (\frac{1}{T_1}- \frac{1}{T_2})

\frac{ In \ \frac{R_1}{ R_2}}{ {\beta}} = (\frac{1}{T_1}- \frac{1}{T_2})

\frac{1}{T_2} =   \frac{1}{T_1}  -          \frac{ In \ \frac{R_1}{ R_2}}{ {\beta}}

{T_2} =  \frac{\beta T_1}{\beta - In (\frac{R_1}{R_2})T}

Replacing our values into the above equation :

{T_2} =  \frac{3650*373}{3650 - In (\frac{20000}{500})373}

{T_2} =  \frac{1361450}{3650 - 3.6888*373}

{T_2} =  \frac{1361450}{3650 - 1375.92}

{T_2} =  \frac{1361450}{2274.08}

{T_2} = 598.68 \ K

{T_2} = 325.68 \ ^0 \ C

Thus, the thermistor temperature = 325.68 \ ^0 \ C

4 0
3 years ago
A student performs Young’s double-slit experiment with slits that are separated by 0.160 mm. She observes the interference patte
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Answer:

Wavelenght is 4.53x10^-7m

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Detailed explanation and calculation is shown in the image below

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