Please need help for this one

Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4

maria [59]

Answer: $247.12\ N$

Explanation:

Given

Magnitude of the charges

$q_1=4.47\times 10^{-6}\ C$

$q_2=1.86\times 10^{-6}\ C$

Distance between them $d=17.4\ mm$

As both charges are of same sign, they must repel each other

Force experienced by second charge is

$\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N$

Thus, charge 2 experience a force of $247.12\ N$

3 0

2 years ago

Read 2 more answers

Technician A says that one of advantages of a clutch brake is its ability to bring a vehicle to a halt at low speeds. Technician

djverab [1.8K]

Answer:

Option B

Explanation:

The advantage of using a clutch brake is that it is capable of handling high load torque and ensures safety at high rotational speeds.

The motivation behind a clutch brake is to stop or slow the information shaft from turning, enabling the apparatuses to work without pounding/conflicting. This kills harm to non-synchronized transmissions, and limits the exertion required when moving from impartial into first or switch from a halt.

7 0

2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

On the basis of which characteristics jute is considerd as a natural fibre

klemol [59]

Answer:

Jute is obtained from Jute plant naturally that is why it is considered as a Natural Fibre. It seems golden in color.

6 0

2 years ago

A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o

Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So $mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}$ .

Kinetic energy of system at first: $\frac{mv^2}{2}=784\;\textbf{J}$ . After: $\frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}$ . The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0

2 years ago