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oksano4ka [1.4K]
2 years ago
9

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-

directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.
Physics
1 answer:
sergey [27]2 years ago
8 0

q = -21 * 10^{-6} C

<h3>What is Free-fall acceleration?</h3>

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\

and its sign must be negative so that it will have upward electric force

so it is

q = -21 * 10^{-6} C

The charge of a particle of mass is -21 * 10^{-6} C

Learn more about Free-fall acceleration: brainly.com/question/2165771

#SPJ1

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If an object is thrown upward at 128 feet per second from a height of 76 feet, its height S after t seconds is given by the foll
faltersainse [42]
<h2>a) Average velocity in first 4 seconds is 64 ft/s upward</h2><h2>b) Average velocity in second 4 seconds is 63.5 ft/s downward</h2>

Explanation:

a) Given S(t) = 76 + 128t − 16t²

    s(0) = 76 + 128 x 0 − 16 x 0² = 76 ft

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    Displacement in 4 seconds = 332 - 76 = 256 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{256}{4}=64ft/s

    Average velocity in first 4 seconds is 64 ft/s upward

a) Given S(t) = 76 + 128t − 16t²

    s(4) = 76 + 128 x 4 − 16 x 4² = 332 ft

    s(8) = 76 + 128 x 8 − 16 x 8² = 78 ft

    Displacement in 4 seconds = 78 - 332 = -254 ft

    Time = 4 - 0 = 4 s

    \texttt{Velocity = }\frac{-254}{4}=-63.5ft/s

    Average velocity in second 4 seconds is 63.5 ft/s downward

3 0
3 years ago
A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh
alekssr [168]

(a) 392 N/m

Hook's law states that:

F=k\Delta x (1)

where

F is the force exerted on the spring

k is the spring constant

\Delta x is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

- The stretching of the spring is

\Delta x=15 cm-10 cm=5 cm=0.05 m

Solving eq.(1) for k, we find the spring constant:

k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N

And so, the stretch of the spring is

\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm

And since the initial lenght of the spring is

x_0 = 10 cm

The final length will be

x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm

8 0
3 years ago
Read 2 more answers
What is the weight of an object with a mass of 6.0 kg on Earth?
gregori [183]
<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>
  • 6 x 9.807
  • = 58.842 N

hope that helps :))

3 0
3 years ago
Read 2 more answers
Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula
RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

m_1 u_1 + m_2 u_2 = (m_1 +m_2)v

where

m_1 = m_2 = m = 10 kg is the mass of each puck

u_1 = 10 m/s is the initial velocity of the 1st puck

u_2 = -20 m/s is the initial velocity of the 2nd puck

v is the final velocity of the two pucks sticking together

Re-arranging the equation and solving for v, we find:

mu_1 + mu_2 = (m+m)v\\u_1 + u_2 = 2v\\v=\frac{u_1+u_2}{2}=\frac{10-20}{2}=-5 m/s

Learn more about momentum:

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8 0
3 years ago
Why are eight electrons (four pairs) surrounding each non-hydrogen atom the optimal electronic arrangement for covalent molecule
rodikova [14]

Eight electrons surrounding each non-hydrogen atom is the optimal electronic arrangement for covalent molecules because it is needed to achieve an octet structure and is necessary to fill both the s and p subshells of electrons.

<h3>What is Covalent bonding?</h3>

This is the type of bonding which involves the sharing of electrons between atoms of an element.

This is done to achieve an octet configuration thereby making them stable and less reactive thereby making it the most appropriate choice.

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3 0
2 years ago
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