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sladkih [1.3K]
2 years ago
12

What is the magnitude of fs on an object lying on a flat surface without moving, on

Physics
1 answer:
Degger [83]2 years ago
7 0

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

  • magnitude of force on the object, F = 10 N
  • angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

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Balance tubes by spacing them equally around the centrifuge and Always balance tubes with other tubes containing a same volume of liquid are right. 
If you don't space them out equally, you will have a lot of broken glass to clean up...trust me. The same thing can happen if you don't have equal amounts of liquid in each tube, but it doesn't have to be exactly the same in every one.
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3 years ago
A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0
1 year ago
What happens if your arrows are too lightly or heavily spined for your bow?
dmitriy555 [2]

Answer:

... If your arrows are too lightly or heavily spined for your bow, the “archer's paradox” resulting in poor arrow flight and loss of accuracy.

7 0
3 years ago
Jack (mass 59.0 kg ) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 47.0 k
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Answer:

Part(A): The magnitude of Jill's final velocity is \bf{6.59~m/s}.

Part(B): The direction is \bf{42.7^{0}} south to east.

Explanation:

Given:

Mass of Jack, m_{1} = 59.0~Kg

Mass of Jill, m_{2} = 47..0~Kg

Initial velocity of Jack, v_{1i} = 8.00~m/s

Initial velocity of Jill, v_{2i} = 0

Final velocity of Jack, v_{1f}  5.00~m/s

The final angle made by Jack after collision, \alpha = 34.0^{0}

Consider that the final velocity of Jill be v_{2f} and it makes an angle of \beta with respect to east, as shown in the figure.

Conservation of momentum of the system along east direction is given by

~~~~&& m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} \cos \alpha + m_{2}v_{2f}^{x}\\&or,& v_{2f}^{x} = \dfrac{m_{1}(v_{1i} - v_{1f} \cos \alpha)}{m_{2}}

where, v_{2f}^{x} is the component of Jill's final velocity along east. The direction of this component will be along east.

Substituting the value, we have

v_{2f}^{x} &=& \dfrac{(59.0~Kg)(8.00~m/s - 5.00 \cos 34.0^{0}~m/s)}{47.0~Kg}\\~~~~~&=& 4.84~m/s

Conservation of momentum of the system along north direction is given by

~~~~&& v_{2f}^{y} + v_{1f} \sin \alpha = 0\\&or,& v_{2f}^{y} = - v_{1f} \sin \alpha = (8.00~m/s) \sin 34^{0} = 4.47~m/s

where, v_{2f}^{y} is the component of Jill's final velocity along north. The direction of this component will be along the opposite to north.

Part(A):

The magnitude of the final velocity of Jill is given by

v_{2f} &=& \sqrt{(v_{2f}^{x})^{2} + (v_{2f}^{y})^{2}}\\~~~~~&=& 6.59~m/s

Part(B):

The direction is given by

\beta &=& \tan^{-1}(\dfrac{4.47~m/s}{4.84~m/s})\\~~~~&=& 42.7^{0}

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cestrela7 [59]
8 meters per second. To find velocity is to divide distance by total time. so 400/50.
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