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3 years ago

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A bowling ball is dropped on a moon from the height of 14 m. The acceleration of gravity on the moon is 4 m/s2. Determine the ti

me it took for the ball to crash at the surface. Round your answer to the nearest whole number.
b. 1 s
c. 4 s
d. 3 s

1 answer:

goblinko [34]3 years ago

7 0

Answer:

4sec

Explanation:

Since we know the initial velocity=0, acceleration = -4; and the displacemnt =14, we use the equation d = vi • t + ½ • a • t2. So, just plug in the values and solve for t in this case it's 4 sec

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A girl pushes a 1.04 kg book across a table with a horizontal applied force 10 points

mr Goodwill [35]

Answer:

Approximately $11.0\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81 \; \rm N \cdot kg^{-1}$ , and that the tabletop is level.)

Explanation:

Weight of the book:

$W = m \cdot g = 1.04 \; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \approx 10.202\; \rm N$ .

If the tabletop is level, the normal force on the book will be equal (in magnitude) to weight of the book. Hence, $F(\text{normal force}) \approx 10.202\; \rm N$ .

As a side note, the $F_N$ and $W$ on this book are not equal- these two forces are equal in size but point in the opposite directions.

When the book is moving, the friction $F(\text{kinetic friction})$ on it will be equal to

$\mu_{\rm k}$ , the coefficient of kinetic friction, times
$F(\text{normal force})$ , the normal force that's acting on it.

That is:

$\begin{aligned}& F(\text{kinetic friction}) \\ &= \mu_{\rm k}\cdot F(\text{normal force})\\ &\approx 0.35 \times 10.202\; \rm N \approx 3.5708\; \rm N\end{aligned}$ .

Friction acts in the opposite direction of the object's motion. The friction here should act in the opposite direction of that $15.0\; \rm N$ applied force. The net force on the book shall be:

$\begin{aligned}& F(\text{net force}) \\ &= 15.0 \; \rm N - F(\text{kinetic friction}) \\& \approx 15.0 - 3.5708\; \rm N \approx 11.429\; \rm N\end{aligned}$ .

Apply Newton's Second Law to find the acceleration of this book:

$\displaystyle a = \frac{F(\text{net force})}{m} \approx \frac{11.429\; \rm N}{1.04\; \rm kg} \approx 11.0\; \rm m \cdot s^{-2}$ .

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2 years ago

A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

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$\eta$ = Efficiency

According to the statement our values are:

$\eta = 0.3$

$T_2 = 210K$

Replacing we have that

$0.3 = 1- \frac{210}{T_1}$

$\frac{210}{T_1} = 0.7$

$T_1 = \frac{210}{0.7}$

$T_1 = 300K$

Therefore the temperature of the heat source is 300K

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malfutka [58]

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

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Therefore, a moment of inertia

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I=0.054

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The period of the oscillations.T = 1.2042s

Read more about the period of the oscillations. brainly.com/question/14394641

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marusya05 [52]

Answer:

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Explanation:

I'd say meters, cause it's the SI unit of length,

which is a Derived Quantity.

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