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3 years ago

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How does the force of gravity between two bodies change when the distance between them doubles? 1. unable to determine; the mass

is needed. 2. doubles 3. halves 4. quadruples 5. remains the same 6. drops to one quarter of its original value?

1 answer:

Rzqust [24]3 years ago

7 0

6. Drop to one quarter of its original value

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5. A 1052 kg truck, starting from rest, reaches a speed of 20.0 m/s in 6.20 s.

BaLLatris [955]

Answer:

a. 3,392.7 N

b. 3,392.7 N

Explanation:

We are given the following information;

Mass of the truck as 1052 kg
initial speed as 0 m/s
Final speed as 20.0 m/s
Time taken as 6.20 s

#a. We are required to calculate the acceleration;

We need to know the formula of getting acceleration;

a = (v-u)/t

Where v is the final velocity, u is the initial velocity

Therefore;

a = (20 m/s - 0 m/s)/6.20s

= 3.225 m/s²

Thus, the average acceleration of the truck is 3.225 m/s²

#b. We are required to calculate the net force on the truck

We need to know that;

According to the second Newton's law of motion, F=ma

Where F is the net force, m is the mass and a is the acceleration.

Therefore;

Net force, F = mass × Acceleration

= 1052 kg × 3.225 m/s²

= 3,392.7 N

Thus, the net force on the truck is 3,392.7 N

4 0

4 years ago

What creates voltage difference in a battery?

Ksenya-84 [330]

<span>Electrolyte is a chemical medium that allows the flow of electrical charge between the cathode and anode.</span>

4 0

4 years ago

Which type of bond is found between the atoms of a molecule?

Brrunno [24]

Answer:

Covalent Bond is found between the atoms of a molecule.

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3 years ago

Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also

ozzi

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$ ,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$

For maximum B, cos(2πft) = 1. Hence,

$B_{max}=\frac{\mu_0\epsilon_0\pi fRV_m}{d}=\frac{4\pi*10^{-7}*8.85*10^{-12}*\pi*60*0.025*160}{4.7*10^{-3}}=1.784*10^{-12}$ T

b) From r = 0 to r = R = 0.025 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fR^2V_m}{rd}$

From r = R = 0.025 m to r = 0.1 m, by Ampere's Law, the equation will be:

$B = \frac{\mu_0\epsilon_0\pi fRV_m}{d}$

The plot is given in the attachment.

5 0

4 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

7 0

4 years ago