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2 years ago

Name at least three physical properties of the bowling ball.

2 answers:

6 0

Hey there!

Here is your answers:

The proper answers to this question is...

"1. The bowling ball is round
2. The bowling ball has 3 holes in the center
3. The bowling ball has a large mass (meaning its heavier than other balls)"

Reason:

Remember that the physical properties is meaning to state the physical features of a bowling ball. For example its heavy, there id three holes on it, and its a round shape or a sphere. When you come across questions like this just state the features of that object in order to get your answer!

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit

oksian1 [2.3K]2 years ago

4 0

Density - Describes how compact the matter is in a substance. ...
Color - Describes how light reflects off a substance.
Luster - Describes how shiny a substance is. ...
Texture - Describes the smoothness/roughness of a substance.
Odor - Describes the smell of a substance.

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A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it

lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0

3 years ago

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

What is the “p” process for elements heavier than iron

tia_tia [17]

Supernova nucleosynthesis is also thought to be responsible for the creation of rarerelements heavier than iron and nickel, in the last few seconds of a type II supernova event.

3 0

3 years ago

The graph below shows the percentage of unstable nuclei remaining of a radioactive isotope.

mafiozo [28]

i have no clue, sorry. jus tryna get points

8 0

2 years ago

A portable x-ray unit has a step-up transformer. The 120 V input is transformed to the 100 kV output needed by the x-ray tube. T

slega [8]

Answer:

$N_s\approx41667 \hspace{3}lo ops$

Explanation:

In an ideal transformer, the ratio of the voltages is proportional to the ratio of the number of turns of the windings. In this way:

$\frac{V_p}{V_s} =\frac{N_p}{N_s} \\\\Where:\\\\V_p=Primary\hspace{3} Voltage\\V_s=V_p=Secondary\hspace{3} Voltage\\N_p=Number\hspace{3} of\hspace{3} Primary\hspace{3} Windings\\N_s=Number\hspace{3} of\hspace{3} Secondary\hspace{3} Windings$

In this case:

$V_p=120V\\V_s=100kV=100000V\\N_p=50$

Therefore, using the previous equation and the data provided, let's solve for $N_s$ :

$N_s=\frac{N_p V_s}{V_p} =\frac{(50)(100000)}{120} =\frac{125000}{3} \approx41667\hspace{3}loo ps$

Hence, the number of loops in the secondary is approximately 41667.

3 0

3 years ago