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Aloiza [94]
3 years ago
8

If a person is moving to the right, the net force acting upon her cannot be balanced

Physics
1 answer:
Svetach [21]3 years ago
5 0

Yeah its True!!  Its unbalanced because the unbalance fore are unequeal in size and opposite in direction.

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Two identical metal spheres A and B are in contact. Both are initially neutral. 1.0× 10 12 electrons are added to sphere A, then
Vilka [71]

Answer:

8 x 10^-8 C on both the spheres.

Explanation:

Number of electrons added to A = 1 x 10^12

As the sphere A and B are identical to each other so the charge is shared equally.

Charge of one electron = 1.6 x 10^-19 C

Charge on A after wards

= number of electrons after wards x charge of one electron

qA = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C

Similarly, the charge on sphere B afterwards

= number of electrons after wards x charge of one electron

qB = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C

8 0
4 years ago
Metal sphere 1 has a positive charge of 7.00 nc . metal sphere 2, which is twice the diameter of sphere 1, is initially uncharge
MariettaO [177]

Answer:

2.33 nC, 4.67 nC

Explanation:

when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:

V_1 = V_2 (1)

Keeping in mind the relationship between charge, voltage and capacitance:

C=\frac{Q}{V}

we can re-write (1) as

\frac{Q_1}{C_1}=\frac{Q_2}{C_2} (2)

where:

Q1, Q2 are the charges on the two spheres

C1, C2 are the capacitances of the two spheres

The capacitance of a sphere is given by

C=4 \pi \epsilon_0 R

where R is the radius of the sphere. Substituting this into (2), we find

\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 R_2} (3)

we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

R_2 = 2R_1

So the eq.(3) becomes

\frac{Q_1}{4 \pi \epsilon_0 R_1}=\frac{Q_2}{4 \pi \epsilon_0 2R_1}

And re-arranging it we find:

Q_2 = 2Q_1

And since we know that the total charge is

Q_1 + Q_2 = 7.00 nC

we find

Q_1 = 2.33 nC\\Q_2 = 4.67 nC

3 0
4 years ago
Which of the following are neutral particles?<br> Protons<br> Neutrons<br> Electrons<br> Atoms
emmasim [6.3K]

Answer:

i believe its an atom

Explanation:

7 0
3 years ago
Read 2 more answers
What can geologist infer from the rounded and eroded shapes of the Appalachian Mountain
Delvig [45]
<span>The answer is D. mountains have been eroding for millions of years. The Appalachian Mountains is an older group of fold mountains. It is much older than the Himalays that do not exibit this rounded shape. As tectonic forces enable mountain building, forces of weathering, erosion and mass wasting are constantly eating away at them. </span>
6 0
3 years ago
Read 2 more answers
A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.50 × 10 4 1.50×104 rad/s to an angular s
KATRIN_1 [288]

Answer:

0.32 s

Explanation:

Initial angular speed: \omega_i = 1.50 \cdot 10^4 rad/s

Final angular speed: \omega_f = 3.35\cdot 10^4 rad/s

Angular rotation: \theta=2.02\cdot 10^4 rad

The angular acceleration of the drill can be found by using the equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

Re-arranging it, we find \alpha, the angular acceleration:

\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}=\frac{(3.35\cdot 10^4 rad/s)^2-(1.50\cdot 10^4 rad/s)^2}{2(2.02\cdot 10^4 rad)}=22,209 rad/s^2

Now we want to know the time t the drill takes to accelerate from

\omega_i =0

to

\omega_f = 6.90\cdot 10^4 rad/s

This can be found by using the equation

\omega_f = \omega_i + \alpha t

where \alpha is the angular acceleration we found previously. Solving for t,

t=\frac{\omega_f - \omega_i}{\alpha}=\frac{22,209 rad/s^2}{6.90\cdot 10^4 rad/s}=0.32 s

5 0
3 years ago
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