The hypotenuse of an isosceles right triangle is 36 units. to the nearest tenth, what is the length of one of the legs

A horizontal rod 0.250 m long is mounted on a balance and carries a current. At the location of the rod a uniform horizontal mag

Karo-lina-s [1.5K]

Answer:

The current is 7.87 A.

Explanation:

Given that,

Length = 0.250 m

Magnetic field $B= 6.10\times10^{-2}\ T$

Magnetic force = 0.120 N

We need to calculate the current

Using formula of magnetic force

$F=BIL$

$I=\dfrac{F}{BL}$

Where, B = magnetic field

I = current

l = length

F = force

Put the value into the formula

$I=\dfrac{0.120}{6.10\times10^{-2}\times0.250}$

$I=7.87\ A$

Hence, The current is 7.87 A.

7 0

4 years ago

Which airplane has more potential energy, a boeing 787 dreamliner having a mass of 2,450kg at 11,500 m above Earth, or a boeing

Readme [11.4K]

Explanation:

For 787 dreamliner airplane,

Mass, $m_1=2450\ kg$

Height, $h_1=11500\ m$

Potential energy of 787 dreamliner airplane is given by :

$E_1=m_1gh_1\\\\E_1=2450\times 9.8\times 11500\\\\E_1=2.76\times 10^8\ J$

For a boeing 747,

Mass, $m_2=3150\ kg$

Height, $h_2=9000 \ m$

:

Potential energy of a boeing 747 airplane is given by

$E_2=m_2gh_2\\\\E_2=3150\times 9.8\times 9000\\\\E_2=2.77\times 10^8\ J$

So, the potential energy of a boeing 747 airplane is more than the 787 dreamliner airplane.

3 0

3 years ago

Determine the magnitude of the acceleration for the speeding up phase.

Arisa [49]

Determine the magnitude of the acceleration for the speeding up phase. Express your answer to two significant figures and include the appropriate units. I've tried 2.8 m/s^2, and 1.1 m/s^2, but they are both wrong. Determine the magnitude of the acceleration for the slowing down phase.

8 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

8 0

3 years ago

Why do some scientists not use controlled experiments?

Leviafan [203]

Observational studies are a prime example. Observational data is more reflective of the real environments that scientists make their inferences to than controlled experiments. The disadvantage of observational studies is that the variability is far greater.

7 0

4 years ago