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2 years ago

Objects that rest have no forces upon them

Physics

1 answer:

Pani-rosa [81]2 years ago

6 0

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

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The charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electro

shusha [124]

Answer:

$1.6\times 10^{-18} C$

Explanation:

The fundamental charge of a single electron is $1.6\times 10^{-19} C$ .

If there are 10 excess electrons, the net charge that would be measured should be 10 times the fundamental charge of a single electron:

$Q=nq_e\\Q= 10\times 1.6\times 10^{-19} C\\Q= 1.6 \times 10^{-18} C$

3 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

Using diagram differentiate between solenoid and a toroid

damaskus [11]

The Toroid is form when you have wound conductor around circular body. In this case you have magnatic field inside the core but you dont have any poles because circular body dont have ends. This can be used where you want minimum flux leakage and dont need magnatic poles. i.e. toroidal inductor, toroidal transformer.

The Solenoid is forn when you wound conductor around body with limb. In this case magnatic field creates two poles N and S. Solenoids have little bit flux leakage. This used where you want magnatic poles and flux leakage is not an issue. i.e. relay, motors, electromagnates.

1 == toroid

2= solenoid

3 0

3 years ago

Which equation relates charge, time, and current?

Ierofanga [76]

Answer:

I = Δq / t

Explanation:

The quantity of electricity i.e charge is related to current and time according to the equation equation:

Q = It

Δq = It

Where:

Q => is the quantity of electricity i.e charge

I => is the current.

t => is the time.

Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:

Δq = It

Divide both side by t

I = Δq / t

6 0

2 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago