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3 years ago

If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?

Physics

1 answer:

Y_Kistochka [10]3 years ago

7 0

Answer:

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

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a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':

Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

$v^{2}=u^{2}+2as$

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

$v=\sqrt{u^{2}+2as}$

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

$v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s$

Also, v=u+at and making t the subject of the formula

$t=\frac {v-u}{a}$

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

$t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s$

Therefore, it needs 7.6 seconds to travel

7 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 1.70 10

anzhelika [568]

Answer:

Explanation:

Given the following :

Speed (V) = speed of 2.30×10^7 m/s

Acceleration (a) = 1.70×10^13 m/s^2

Using the right hand rule provided by Lorentz law:

B = F / qvSinΘ

Where B = magnitude of the magnetic field

v = speed of the particle

Θ = 90° (perpendicular to the field)

q = charge of the particle

SinΘ = sin90° = 1

Note F = ma

Therefore,

B = ma / qvSinΘ

Mass of proton = 1.67 × 10^-27

Charge = 1.6 × 10^-19 C

B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1

B = 2.839 × 10^-14 / 3.68 × 10^-12

B = 0.7715 × 10^-2

B = 7.72 × 10^-3 T

2) Magnetic field will be in the negative y direction according to the right hand thumb rule.

Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.

5 0

3 years ago

If a car driving down the road doubles its speed what happens to its kinetic energy?

ivanzaharov [21]

Answer:

Kinetic energy doubles

7 0

2 years ago

7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca

Misha Larkins [42]

Answer:

q = C V charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 * 10E-4 C per capacitor

N = Q / q = 1 / 1.1 * 10E-4 = 9091 capacitors

8 0

3 years ago