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2 years ago

A speed-time graph is a horizontal line with a y-value of 4. Which describes the objects motion?

Physics

1 answer:

taurus [48]2 years ago

5 0

It means the speed is constant with a value of 4 units.

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A 10 ohm resistor has 5 a current in it. what is the voltage across the resistor?

Artemon [7]

R = 10 ohm
I = 5 A

V= IR = 10 x 5 = 50 v

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2 years ago

Boron is one position to the left of carbon on the periodic table. The atomic number of carbon is 6. Given its position on the p

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Answer:

Explanation:

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2 years ago

The terminals of a 0.70 Vwatch battery are connected by a 80.0-m-long gold wire with a diameter of 0.200 mm What is the current

Komok [63]

Answer:

$I=0.047A$

Explanation:

Let's use Ohm's law:

$V=IR$

$I=\frac{V}{R}$ (1)

Where:

$V=Voltage\\I=Current\\R=Electrical\hspace{2 mm}Resistance$

We know the value of the voltage V, so we need to find the value of R in order to find I. Fortunately there is a relation between the resistivity of a conductor and its electrical resistance given by:

$R=\rho*\frac{l}{A}$ (2)

Where:

$R=Electrical\hspace{2 mm}Resistance\\l=Length\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=80m\\A=Cross\hspace{2 mm}sectional\hspace{2 mm}area\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}conductor=1.256637061*10^{-7} \\\rho=Electrical\hspace{2 mm}resistivity\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}material=2.35*10^{-8}$

Keep in mind that the electrical resistivity of the gold is a known constant which is $\rho_g_o_l_d=2.35*10^{-8}$ and the cross sectional area of the conductor is calculated as:

$A=\pi *(r^{2})=\pi *(0.0002m)^{2} =1.256637061*10^{-7} m^{2}$

Because we have a wire in this case, so we assume a cylindrical geometry.

Now replacing our data in (2)

$R=(2.35*10^{-8})*\frac{80}{1.256637061*10^{-7} } =14.96056465\Omega$

Finally, we know R and V, so replacing these values in (1) we will be able to find the current:

$I=\frac{0.7}{14.96056465}\approx0.047A$

7 0

2 years ago

Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

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2 years ago

A glass of root beer with a scoop of ice cream floating on top and a straw sticking out.

vampirchik [111]

Answer:

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1 year ago