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2 years ago

15

If the mass of the cart was increased but the hanging mass remained the same, how would the acceleration be affected? Explain ho

w you know in terms of net force and system mass.

1 answer:

Margaret [11]2 years ago

6 0

Answer:

It is always said that mass is same everywhere.

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The average mass of an automobile in the United States is about 1.440x10^6 g express this mass in kilograms

-BARSIC- [3]

From the problem statement, this is a conversion problem. We are asked to convert from units of grams to units of kilograms. To do this, we need a conversion factor which would relate the different units involved. We either multiply or divide this certain value to the original measurement depending on what is asked. From literature, we will find that 1000 grams is equal to 1 kilogram. We use this as follows:

<span> 1.440x10^6 g ( 1 kg / 1000 g ) = 1440 kg</span><span>
</span>

8 0

2 years ago

Consider two circular metal wire loops each carrying the same current I as shown below. In what r... Consider two circular metal

NeX [460]

Answer:

1) The magnetic field outside the loop is zero.

In region III the magnetic fields due to the two wire loops point in the opposite direction andhence cancel each other. Therefore the magnetic field is zero in region I, III and V

The diagram is attached

6 0

2 years ago

The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star

Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

$PV=nRT$

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3$

V=8.86*10^{50}L

and for n

$n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol$

Hence, we have (1 Pa = 9.85*10^{-9}atm)

$T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K$

hope this helps!!

8 0

2 years ago

A typical garden hose has an inner diameter of 5/8". Let's say you connect it to a faucet and the water comes out of the hose wi

castortr0y [4]

Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
$v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \: \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142$
$we \: can \: only \: assume \:that \\ flow \: (q) \:stays \: same \: since \: it \\ isnt \: impeded \: by \: anything \\ thus \:it \: (q)\: stays \: the \: same \: \\ so \: 4q \: can \: be \: removed \: from \: \\ the \: equation$
$then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to > > \\ (v1 \times {d1}^{2} \pi) = (v2 \times {d2}^{2}\pi)$
$both \: \pi \: will \: cancel \: each \: other \: out \: \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \: =$

$(v1 \times {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times {0.052}^{2}) = (v2 \times {0.021}^{2}) \\ divide \: both \: sides \: by \: {0.021}^{2} \\ to \: solve \: for \: v2 > >$
$v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2}) \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second$
new velocity coming out of the hose then is
44 ft/sec

4 0

2 years ago

Which of the following most accurately represents John Dalton’s model of the atom? A. a tiny, solid sphere with an unpredictable

aleksley [76]

A and c are the answersss

6 0

2 years ago