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3 years ago

Can someone solve this problem and explain to me how you got it

Physics

1 answer:

evablogger [386]3 years ago

4 0

Answer:

question5: F=74312.5N

question6: charge at the end of antenna=0.37N

Explanation:

Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.

⇒ $F\alpha\frac{q1*q2}{r^{2}}$

∴ $F=k\frac{q1*q2}{r^{2}}$

where $F$ is the force of attraction or repulsion

$k$ is Coulumb's constant= $9*10^{9}Nm^{2}C^{-2}$

$q1$ and $q2$ are the magnitude of the charges

$r$ is the distance between two charges

The force between the two charges is attractive if they are of different polarity

The force between the two charges is repulsive if they are of same polarity

Question5:

Given: q1=0.041 C, q2=0.029 C, r=12 m

therefore by Coulumb's law,

$F=9*10^{9}*\frac{0.041*0.029}{12^{2}}$

$F=74312.5N$

Question6:

Given: q1= $3*10^{-18}C$ , r=5 m, F= $4*10^{-11}N$

therefore by Coulumb's law,

$4*10^{-11}=9*10^{9}*\frac{3*10^{-18}*q2}{5^{2}}$

⇒ $q2=\frac{4*10^{-11}*25}{9*10^{9}*3*10^{-18}} \\=0.37C$

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A person wants to lose weight by "pumping iron". The person lifts an 80 kg weight 1 meter. How many times must this weight be li

statuscvo [17]

Answer:

37357 sec

or 622 min

or 10.4 hrs

Explanation:

GIVEN DATA:

Lifting weight 80 kg

1 cal = 4184 J

from information given in question we have

one lb fat consist of 3500 calories = 3500 x 4184 J

= 14.644 x 10^6 J

Energy burns in 1 lift = m g h

= 80 x 9.8 x 1 = 784 J

lifts required $= \frac{(14.644 x 10^6)}{784}$

= 18679

from the question,

1 lift in 2 sec.

so, total time = 18679 x 2 = 37357 sec

or 622 min

or 10.4 hrs

3 0

3 years ago

When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What

yarga [219]

Answer:

$\Delta E=1.64*10^{-18}J$

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

$\Delta E=h\nu(1)$

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

$\nu=\frac{c}{\lambda}(2)$

Here c is the speed of light. Replacing (2) in (1):

$\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J$

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3 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

3 years ago

A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.

Crank

Answer: 1000 Hz

Explanation:

You can calculate frequency by dividing velocity by wavelength

Frequency = velocity/wavelength

Find velocity first.

900 m/3 s = 300 m/s

Plug values in to find frequency.

F = (300 m/s)/0.3 m

F = 1000 Hz

8 0

3 years ago

A rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of

choli [55]

Answer:

Explanation:

Let f be the amount of fuel oxidizer needed

v be the speed

The relationship between them is inverse in nature i.e

f ∝ 1/v

f = k/v

If a rocket for use in deep space is to have the capability of boosting a total load (payload plus the rocket frame and engine) of 3.25 metric tons to a speed of 10,000 m/s, then f = 3.25 when v = 10,000

Substitute and get k

k = fv

k = 3.25 * 10,000

k = 32500

To get the amount of fuel oxidizer required to produce a speed of 2400m/s, we will find f when v = 2400m/s

Recall that f = k/v

f = 32500/2400

f = 13.54 metric tons

<em>Hence the fuel plus oxidizer that will be required is 13.54 tons</em>

4 0

3 years ago

Can someone solve this problem and explain to me how you got it​

Can someone solve this problem and explain to me how you got it