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2 years ago

A 1.5 V battery produces 41.8 mA when it is across a 30.0 Ω load. The potential difference across the battery's terminals is

Physics

2 answers:

Airida [17]2 years ago

6 0

Answer:

Explanation:

V = I R

V = 0.0418 * 30

V = 1.254

this battery seems dead :P

olchik [2.2K]2 years ago

5 0

Answer:

Explanation:

Let's just assume that the battery has no internal resistance. If we get an absurd answer, we'll come back and look at that assumption.

E = I * R

I = 41.8/1000 amps = 0.0418

R = 30.0 ohms

E = 0.0418 * 30

E = 1.254 volts. This is an OK answer. The battery is a little old.

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Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

Daniel [21]

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

$v_f^2 - v_i^2 = 2 ad$

so here we have

$0^2 - 40^2 = 2 a (70 feet)$

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

$0^2 - 48^2 = 2 a (d)$

now divide the above two equations

$\frac{40^2}{48^2} = \frac{70 feet}{d}$

$d = 100.8 ft$

4 0

3 years ago

A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have

DiKsa [7]

Answer:

4.4×10⁻⁷ Coulomb

Explanation:

V = Voltage = 5.8 kV

d = Potential distance = 2.8 mm = 0.0028 m

A = Area = 0.3×0.08 = 0.024 m²

ε₀ = permittivity constant in a Vacuum= 8.85×10⁻¹² F/m

$\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

6 0

3 years ago

Please help me with this physics prooblem

zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

$x=v_0\cos20.0^\circ t+\dfrac12a_xt^2$

$y=v_0\sin20.0^\circ t+\dfrac12a_yt^2$

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

$x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ$

$y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ$

So we have enough information to solve for the components of the acceleration vector, $a_x$ and $a_y$ :

$x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}$

$y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}$

The acceleration vector then has direction $\theta$ where

$\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ$

5 0

2 years ago

(b) Which has more mass: 50 cm of gold or 50 cm of aluminum? Explain.

monitta

Gold’s molar mass is about 196 while aluminum is about 27, thus 50cm of gold has more mass

8 0

3 years ago

The data table shows some data related to the Sun and the planets in our solar system.

AlexFokin [52]

You have selected the correct answer and blobbed over it with your pencil.

I assume you must have looked at Saturn's average distance, found 1427,
divided that number by 6, got 237 and change, then looked at the others,
and found that 228 was the only one that's anywhere close.

8 0

3 years ago

Read 2 more answers