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2 years ago

15

Martina has a sample of an unknown substance. She measures the substance. Its mass is 13.5 grams, and its volume is 5 cm3. Which

metal might her sample be made of?

1 answer:

yaroslaw [1]2 years ago

3 0

Answer:

Aluminum Density

Explanation:

2.7 g/cm3

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A child tugs on a rope attached to a 0.62-kg toy with a horizontal force

Katarina [22]

Answer:

$a=0.8\ m/s^2$

Explanation:

Given that,

Mass of a toy, m = 0.62 kg

Horizontal force, F = 16.3 N

Force on the opposite direction, F' = 15.8 N

We need to find the acceleration of the toy.

Here, two forces are acting in opposite direction, the net force will be the difference of forces.

Net force = 16.3 N-15.8 N

=0.5 N

The formula for net force is given by :

F = ma

a is the acceleration of the toy

$a=\dfrac{F}{m}\\\\a=\dfrac{0.5\ N}{0.62\ kg}\\\\a=0.8\ m/s^2$

So, the acceleration of the toy is $0.8\ m/s^2$ .

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2 years ago

If you add 700 kJ of heat to 700 g of water at 70 degrees C, how much water is left in the container? The latent heat of vaporiz

makkiz [27]

Answer:A

Explanation:Find attached picture file for details

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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

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2 years ago

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

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2 years ago

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A small object carrying a charge of -3.00 nc is acted upon by a downward force of 30.0 nn when placed at a certain point in an e

arlik [135]

The working equation for this one is:

E = F/Q, where E is the strength of the electric field, F is the electric force and Q is the charge. Substituting the corresponding values, the strength of the electric field is equal to

E = -30 nN/-3 nC
E = 10 nN/nC

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2 years ago