When a rubber ball is thrown against a wall, energy is transferred and transformed. Select the

Which of the following ways of writing 1000w is incorrect?

xenn [34]

Answer:

the third one is incorrect

Explanation:

10 x 10³= 10^1 x 10^3 = 10^4

8 0

3 years ago

a swimmer experiences a total (absolute) pressure of 117,500 pa in a pool. how far below the surface are they located?

marta [7]

Answer:

Explanation:

We know that the pressure can be calculated in the following way:

p = d·g·h

with d being the density of the water, g the gravitational acceleration and h the depth.

Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa

117,500 Pa = 1000kg/m³ · 9.8m/s² · h

Therefore h = 11,9 m

4 0

2 years ago

Please help... I'm confused on what I represents in terms of solving the total current. Would variable would I be singling out?

pentagon [3]

Answer:

the researcher say hi for us the best pa the best of us are going out to eat that I can get my money toward a little bit but the best of luck to be at work by then and we will see what the status

7 0

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

2 years ago

The mass of the Earth is 6 × 1024 kg, the mass of the Moon is 7 × 1022 kg, and the center-to-center distance is 4 × 108 m. How f

Karolina [17]

Answer:

The center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

Explanation:

Let the reference point be the center of the Earth

$X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m}$

Where;

Xcm is the distance from center of the Earth =?

Me is the mass of the Earth = 6 × 10²⁴ kg

Xe is the center mass of the Earth = 0

Mm is the mass of the moon = 7 × 10²² kg

Xm is the center mass of the moon = 4 × 10⁸ m

$X_{Cm = \frac{M_eX_e +M_mX_m}{M_e+M_m} = \frac{M_e(0) +M_mX_m}{M_e+M_m} = \frac{ M_mX_m}{M_e+M_m}}\\\\X_C_m = \frac{7 X 10^{22}*4X10^8}{6X10^{24}+7X10^{22}} =\frac{28 X10^{30}}{607X10^{22}}\\\\ X_C_m = 4.613 X10^6 m$

Therefore, the center of mass of the Earth–Moon system is 4.613 × 10⁶ m from center of the Earth.

8 0

4 years ago