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Lubov Fominskaja [6]

3 years ago

15

A car that experiences no frictional force is started and caused to move. For the car to continue in that motion, the gas pedal

would have to be used:

1 answer:

astraxan [27]3 years ago

8 0

Explanation:

Gas pedal is not required to used on friction less surface. On a friction less surface once the car is started and caused to move, car continues to move with same velocity as there is no opposing force. Therefore, no pressing of gas pedal required. However, this situation is not possible in real life, as friction is always present.

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Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called ____.

irina1246 [14]

Answer:

Small, icy bodies that have highly eccentric orbits and can be found in the Oort cloud or the Kuiper belt are called COMETS.

4 0

3 years ago

A physical pendulum in the form of a planar object moves in simple harmonic motion with a frequency of 0.460 Hz. The pendulum ha

zlopas [31]

Answer:

The moment of inertia is $I =1.0697 \ kg m^2$

Explanation:

From the question we are told that

The frequency is $f = 0.460 \ Hz$

The mass of the pendulum is $m = 2.40 \ kg$

The location of the pivot from the center is $d = 0.380 \ m$

Generally the period of the simple harmonic motion is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{I}{ m * g * d } }$

Where I is the moment of inertia about the pivot point , so making I the subject of the formula it

=> $I = [ \frac{T}{2 \pi } ]^2 * m* g * d$

But the period of this simple harmonic motion can also be represented mathematically as

$T = \frac{1}{f}$

substituting values

$T = \frac{1}{0.460}$

$T = 2.174 \ s$

So

$I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380$

$I =1.0697 \ kg m^2$

4 0

3 years ago

If v=10v, what value of R would limit l to 10 mA? (if you have 10 volts, what resistance would be needed to limit the current to

777dan777 [17]

r=v/i =10/0.01=1000ohms B

7 0

3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

A +12 Î¼C charge and -8 Î¼C charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t

Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle $q_1=+12\ \mu C$

Charge of second Particle $q_2=-8\ \mu C$

distance between them $d=4\ cm$

$k=9\times 10^{9}$

magnetic field due to first charge at mid-way between two charged particles is

$E_1=\frac{kq_1}{r^2}$

$r=\frac{d}{2}=\frac{4}{2}=2\ cm$

$E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_1=27\times 10^7\ N/C$ (away from it)

Electric field due to $q_2=-8\ \mu C$

$E_2=\frac{kq_2}{r^2}$

$E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}$

$E_2=-18\times 10^7\ N/C$ (towards it)

$E_{net}=E_1+E_2$

$E_{net}=9\times 10^7\ N/C$ (away from first charge)

8 0

3 years ago