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3 years ago

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A bowling ball rolls 33\,\text m33m33, start text, m, end text with an average speed of 2.5\,\dfrac{\text m}{\text s}2.5 s m 2

, point, 5, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. How many seconds did the ball roll?

1 answer:

asambeis [7]3 years ago

4 0

The ball rolled for 13.2 s

<h3>Further explanation</h3>

Speed is scalar and no direction

$\tt avg~speed=\dfrac{total~distance}{elapsed~time}=\dfrac{\Delta x}{\Delta t}$

A bowling ball rolls 33 m, with average speed = 2.5 m/s

So elapsed time :

$\tt t=\dfrac{33~m}{2.5`m/s}=13.2~s$

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A 0.55 kg projectile is launched from the edge of a cliff with an initial kinetic energy of 1550 J and at its highest point is 1

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Answer:

a). 53.78 m/s

b) 52.38 m/s

c) -75.58 m

Explanation:

See attachment for calculation

In the c part, The negative distance is telling us that the project went below the lunch point.

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3 years ago

A school bus takes 0.530 h to reach the school from your house. If the average velocity of the bus is 19.0 km/h to the east, wha

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Answer:

$x_{distance}=10.07km$

Explanation:

Given data

time=0.530 h

Average velocity Vavg=19.0 km/s

To find

Displacement Δx

Solution

The Formula for average velocity is given as

$V_{avg}=(x_{distance} )/(t_{time} )\\ x_{distance}=V_{avg}*(t_{time} )\\x_{distance}=(19.0km/h)*(0.530h)\\x_{distance}=10.07km$

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4 years ago

How much pressure is on the bottom of a pot that holds 20N of soup? the surface area of the pot is 0.05m2

fomenos

Answer:

400

Explanation:

Formula used in solution:

$P = \frac{F}{A}$

$P - pressure,\: F - force,\: A - area$

The given information:

$F = 20N$

$A = 0.05m^{2}$

Solution

$P = \frac{F}{A} = \frac{20}{0.05} = 20 * 20 = 400$

Answer:

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3 years ago

PROBLEM 5 (Problem 4-145 in 7th edition) Consider a well-insulated horizontal rigid cylinder that is divided into two compartmen

kari74 [83]

Answer:

The answer is " $\bold{83.8^{\circ} \ C}$ ".

Explanation:

Formula for calculating the mass in He:

$\to m = \frac{PV}{RT}\\$

$= \frac{500 \times 1}{ 2.0769 \times (40 + 273)}\\\\ = \frac{500 }{ 2.0769 \times 313}\\\\ = \frac{500 }{ 650.0697}\\\\= 0.76914 \ Kg$

Formula for calculating the mass in $N_2$ :

$\to m = \frac{PV}{RT}\\$

$= \frac{500 \times 1}{ 0.2968 \times (120+ 273)}\\\\ = \frac{500 }{ 0.2968 \times 393}\\\\ = \frac{500 }{ 116.6424}\\\\= 4.2866\ Kg$

by using the temperature balancing the equation:

$T' = \frac{mcT (He) + mcT ( N_2 )}{ mc (He) + mc ( N_2)}$

$= \frac{0.76914 \times 3.1156 \times 313 + 4.2866 \times 0.743 \times393}{ 0.76914 \times 3.1156 + 4.2866 \times 0.743} \\\\ = 357 \ \ K \approx 83.8^{\circ} \ C$

3 0

3 years ago

Ball A 1.55kg moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B m

Illusion [34]

1.15 m/s to the left (3 sig. fig.).

<h3>Explanation</h3>

Momentum is conserved between the two balls if they are not in contact with any other object. In other words,

$p_{\text{A,initial}} + p_{\text{B,initial}}=p_{\text{A,final}} + p_{\text{B,final}}$

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ , where

$m$ stands for mass and
$v$ stands for velocity, which can take negative values.

Let the velocity of objects moving to the right be positive.

$m_\text{A} = 1.55\;\text{kg}$ ,
$m_\text{B} = 0.752\;\text{kg}$ .

Before the two balls collide:

$v_\text{A} = +8.76\;\text{m}\cdot\text{s}^{-1}$ ,
$v_\text{B} = -11.4\;\text{m}\cdot\text{s}^{-1}$ .

After the two balls collide:

$v_\text{A}$ needs to be found,
$v_\text{B} = +9.03\;\text{m}\cdot\text{s}^{-1}$ .

Again,

$m_\text{A} \cdot v_{\text{A,initial}} + m_\text{B}\cdot v_{\text{B,initial}}=m_\text{A}\cdot v_{\text{A,final}} + m_\text{B}\cdot v_{\text{B,final}}$ ,

$1.55 \times (+8.76) + 0.752 \times (-11.4) = 1.55\;{\bf v_{\textbf{A,final}}} + 0.752 \times (+9.03)$ .

$v_{\text{A,final}} = \dfrac{1.55 \times (+8.76) + 0.752 \times (-11.4)-0.752 \times (+9.03)}{1.55} = -1.15\;\text{m}\cdot\text{s}^{-1}$ .

$v_{\text{A,final}}$ is negative? Don't panic. Recall that velocities to the right is considered positive. Accordingly, negative velocities are directed to the left.

Hence, ball A will be travelling to the left at 1.15 m/s (3 sig. fig. as in the question) after the collision.

8 0

3 years ago

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