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1 year ago

At a distance r from a charge e on a particle of mass m the electric field value is

1 answer:

7 0

At a distance r from a charge e on a particle of mass m the electric field value is 8.9876 × 10⁹ N·m²/C². Divide the magnitude of the charge by the square of the distance of the charge from the point. Multiply the value from step 1 with Coulomb's constant.

<h3>what is magnitude ?</h3>

Magnitude can be defined as the maximum extent of size and the direction of an object.

It is used as a common factor in vector and scalar quantities, as we know scalar quantities are those quantities that have magnitude only and vector quantities are those quantities have both magnitude and direction.

There are different ways where magnitude is used Magnitude of earthquake, charge on an electron, force, displacement, Magnitude of gravitational force

For more details regarding magnitude, visit

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What is the theory of light?

Lostsunrise [7]

Answer:

Corpuscular theory of light

Explanation:

In optics, the corpuscular theory of light, arguably set forward by Descartes in 1637, states that light is made up of small discrete particles called "corpuscles" which travel in a straight line with a finite velocity and possess impetus. This was based on an alternate description of atomism of the time period.

3 0

3 years ago

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th

jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

R = Vo² sin (2θ) / g

sin 2θ = g R / Vo²

sin 2θ = 9.8 75/35²

2θ = sin⁻¹ (0.6)

θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

X = Vox t

t2 = X2 / Vox = X2 / (Vo cosθ)

t2 = 37.5 / (35 cos 18.4)

t2 = 1.13 s

With this time we calculate the height at this point

Y = Voy t - ½ g t²

Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²

Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

8 0

3 years ago

Read 2 more answers

A track star runs 100 meters in 10 seconds. What is the star's average speed?

kakasveta [241]

Answer:

10m/s

Explanation:

4 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

Which three quantities can be used to calculate acceleration?

PtichkaEL [24]

D is the correct answer, assuming that this is the special case of classical kinematics at constant acceleration. You can use the equation V = Vo + at, where Vo is the initial velocity, V is the final velocity, and t is the time elapsed. In D, all three of these values are given, so you simply solve for a, the acceleration.
A and C are clearly incorrect, as mass and force (in terms of projectile motion) have no effect on an object's motion. B is incorrect because it is not useful to know the position or distance traveled, unless it will help you find displacement. Even then, you would not have enough information to use a kinematics equation to find a.

4 0

4 years ago