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3 years ago

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An object with a charge of -2.9 μC and a mass of 1.0×10^−2 kg experiences an upward electric force, due to a uniform electric fi

eld, equal in magnitude to its weight. Find the magnitude of the electric field?
If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.?

1 answer:

ki77a [65]3 years ago

8 0

Answer:

$E=3.38*10^{4}N/C$
The new acceleration will have a value equal to gravity (9.81m/s^2) but pointing upwards.

Explanation:

The electric Force and gravitational Force are the same:

$m*g=E*q$

$E=m*g/q=1.0*10^{-2}*9.81/(2.9*10^{-6})=3.38*10^{4}N/C$

If the Electric field is doubled, the Electric Force is doubled, so its new value is twice the weight of the object. If we add this new electric force (upwards), with the weight (downwards), we have a resulting force upwards = one time the weight. in conclusion the acceleration will have a value equal to gravity but pointing upwards.

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Marina CMI [18]

Answer:

Crumple zones are designed to absorb and redistribute the force of a collision. ... Also known as a crush zone, crumple zones are areas of a vehicle that are designed to deform and crumple in a collision. This absorbs some of the energy of the impact, preventing it from being transmitted to the occupants.

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3 years ago

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

$16 \ m/s = v_i + 12 \ m/s$

We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

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The cyclist's initial velocity is <u>4 meters per second.</u>

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3 years ago

Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char

lorasvet [3.4K]

Answer:

$F'=2F$

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

$F=\frac{kq_1q_2}{d^2}$

In this case, we have $q_1'=2q_1$ :

$F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F$

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3 years ago

An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t

Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

$A_{1}V_{1}=A_{2}V_{2}$

Put the value into the formula

$16.0\times3.8\times2.8=16.5\times x\times 11.0$

$x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}$

$x=0.938\ m$

Hence, The depth of the water at this point is 0.938 m.

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4 years ago

A horizontal uniform board of weight 125N and length 4 m is supported by vertical chains at each end. A person weighing 500N sit

Misha Larkins [42]

Let's apply an equation of equilibrium to the situation: The sum of the moments about the left end of the board must equal 0.

We have three moments to add. Positive force values indicate upward direction and negative values indicate downward direction. All distances given below are measured to the right side of the left end of the board:

The weight of the board, -125N, located at 2m (center of the board due to its uniform density)
The tension in the right chain, +250N, located at 4m
The weight of the person, -500N, located at a distance "x"

The sum of the moments must equal 0 and is given by:

ΣFx = 0

F is the magnitude of force, x = distance from the left end of the board

Plug in all of the force and distance values and solve for x:

ΣFx = 250(4) - 125(2) - 500x = 0

500x = 750

x = 1.5m

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3 years ago