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Serggg [28]
3 years ago
9

A battery supplies E volts to a circuit containing a resistor R. How could the current in the resistor be increased? Select all

that apply.
Replace the resistor with a smaller resistor.
Replace the resistor with a larger resistor.
Connect a smaller battery.
Connect a larger battery.
Physics
1 answer:
anastassius [24]3 years ago
5 0

Answer:

I=V/R

so larger bsttery or small resistor can increase the current value.

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Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728
Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19  

= 9*1.007 + 10*1.008 = 19.152 u

Actual  mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23  

= 1.38*10^13 J/mol

8 0
3 years ago
Consider electrons of kinetic energy 6.0 eV and 600 keV. For each electron, find the de Broglie wavelength, particle speed, phas
irinina [24]

Answer:

For 6.0 eV

0.5 nm, 1.45*10^6 m/s, 6.17*10^10 m/s, 1.45*10^6 m/s

For 600 eV

1.26*10^-3 nm, 2.66*10^8 m/s, 3.37*10^8 m/s, 2.66*10^8 m/s

Explanation:

See attachment for calculation

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3 years ago
You are trying to catch the mutated mouse and you have a rope
Wittaler [7]

Answer:

none no work cuz no motion

Explanation:

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2 years ago
The purpose of the air filter is to
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Read 2 more answers
The drive chain in a bicycle is applying a torque of 0.850 nm to the wheel of the bicycle. treat the wheel as a hoop with a mass
BlackZzzverrR [31]
The equivalent of Newton's second law for rotational objects is given by:
\tau = I \alpha
where
\tau is the net torque acting on the object
I is its moment of inertia
\alpha is its angular acceleration

For a hoop rotating around its perpendicular axis, the moment of inertia is
I=mr^2
where m is the mass and r the radius. By using the data of the wheel, m=0.750 kg and r=33.0 cm=0.33 m, we find
I=mr^2 = (0.750 kg)(0.33 m)^2=0.082 kg m^2

and since the torque is \tau=0.850 Nm, the angular acceleration of the wheel is
\alpha= \frac{\tau}{I}= \frac{0.850 Nm}{0.082 km m^2}=10.37 rad/s^2
6 0
3 years ago
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