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3 years ago

8

one ball rolls along a shelf at a steady rate. a second ball rolls off the shelf and gains speed as it falls in a curved path. w

hich must have an unbalanced force acting on it?

1 answer:

Salsk061 [2.6K]3 years ago

6 0

The ball that rolls at a steady rate

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Bicarbonato es metal o no?

givi [52]

Answer:

La palabra bicarbonato es un término químico para referirse a una sal ácida del ácido carbónico en combinación con un metal. ... Pero si sólo está sustituido un átomo de hidrógeno por el metal, entonces obtenemos un bicarbonato o carbonato ácido, por ejemplo, bicarbonato sódico: CO3H Na.

7 0

3 years ago

Must show units and how they cancelli 1.) 175 km to um 3.) 385 nm to dm 5.) 492 um tom 7.) 52 x 103 dm to mm 9.) 321x 1035 mm to

morpeh [17]

Explanation:

1.) 175 km to μm

$1 km=10^9 \mu m$

$175 km=175\times 10^9\mu m=1.75\times 10^{11} \mu m$

3.) 385 nm to dm

$1 nm=10^{-8} dm$

$385 nm=385\times 10^{-8} dm=3.85\times 10^{-6} dm$

5.) 492 μm to m

1 μm = $10^{-6} m$

$492 \μm=492\times 10^{-6} m=4.92\times 10^{-4} m$

7.) $52\times 10^3$ dm to mm

1 dm = 100 mm

$52\times 10^3 dm=52\times 10^3\times 100 mm=5.2\times 10^{6}dm$

9.) $321\times 10^{35}$ mm to km

$1 mm = 10^{-6} km$

$321\times 10^{35} mm=321\times 10^{35}\times 10^{-6} km=3.21\times 10^{31} km$

11.) $456\times 10^3$ m to km

m = 0.001 km

$456\times 10^3m =456\times 10^3 m\times 0.001 km=456 km$

13.) $422\times 10^3$ m to nm

$1 m = 10^{9} nm$

$422\times 10^3 m=422\times 10^3\times 10^{9} nm=4.22\times 10^{14} nm$

15.) $4.87\times 10^{30}$ m to pm

$1 m = 10^{12} pm$

$4.87\times 10^{30} m=4.87\times 10^{30}\times 10^{12} pm=4.82\times 10^{42} pm$

17.) $5.26\times 10^3$ m to um

1 m = $10^{6} \mu m$

$5.26\times 10^3 m=5.26\times 10^3\times 10^6 \mu m=5.26\times 10^{9} \mu m$

19.) $1.25\times 10^{35}$ m to Mm

1 m = $10^{-6} Mm$

$1.25\times 10^{35} m=1.25\times 10^{35}\times 10^{-6} Mm=1.25\times 10^{-29} Mm$

21.) $4.22\times 10^3$ Tm to nm

$1 Tm = 10^{21} nm$

$4.22\times 10^3 Tm=4.22\times 10^3\times 10^{21} nm=4.22\times 10^{24} nm$

6 0

3 years ago

During which phase of mitosis does the cell begin to split into two daughter cells

artcher [175]

Answer:

$\text{Telophase}$

Explanation:

<em>Telophase is the process that separates the duplicated genetic material carried in the nucleus of a parent cell into two identical daughter cells.</em>

<em />

<em /> $-TheAnimeGirl-$ <em />

3 0

2 years ago

Read 2 more answers

How many moles of oxygen gas, o2, are in a storage tank with a volume of 1.000×105 l at stp?

Mumz [18]

STP means standard temperature and pressure which is equivalent to 273 K and 1 atm, respectively. Assuming ideal gas behavior, the solution for this problem is as follows:

PV = nRT
Solve for n,
n = RT/PV
n = (0.0821 L-atm/mol-K)(273 K)/(1 atm)(1×10⁵ L)
<em>n = 2.24×10⁻⁴ moles</em>

6 0

3 years ago

Read 2 more answers

What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)

Alexxandr [17]

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.

Substitute these values into equation 4

V = (301×18.4×0.082)/(32×1.3132)

V = 454.1488/42.0224

V = 10.81 dm³

V = 11.0 dm³

8 0

3 years ago