A 2-C charge experiences a force of 40 N when put at a certain location inspace. The electric field at that location is a. 2 N/C.b. 20 N/C. c. 30 N/C. d.
40 N/C. e. 60 N/C.

Answers

Answer 1
Answer:

Answer:

E = 20 N/C

Explanation:

Given that,

Charge, q = 2 C

Force experience, F = 40 N

We need to find the electric field at that location.

The electric field in terms of electric force is given by :

F = qE

Where

E is the electric field

E=(F)/(q)\n\nE=(40\ N)/(2\ C)\nE=20\ N/C

So, the electric field at that location is 20 N/C.


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During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 176 N on the wire. The left section of the wire makes an angle of 12.5° relative to the horizontal and sustains a tension of 413 N. Find the (a) magnitude and (b) direction (as an angle relative to horizontal) of the tension that the right section of the wire sustains.
A uniform, solid sphere of radius 3.75 cm and mass 4.00 kg starts with a purely translational speed of 1.75 m/s at the top of an inclined plane. The surface of the incline is 3.00 m long, and is tilted at an angle of 26.0∘ with respect to the horizontal. Assuming the sphere rolls without slipping down the incline, calculate the sphere's final translational speed ????2 at the bottom of the ramp.

A spinning wheel on a fireworks display is initially rotating in a counterclockwise direction. The wheel has an angular acceleration of -4.46 rad/s2. Because of this acceleration, the angular velocity of the wheel changes from its initial value to a final value of -31.4 rad/s. While this change occurs, the angular displacement of the wheel is zero. (Note the similarity to that of a ball being thrown vertically upward, coming to a momentary halt, and then falling downward to its initial position.) Find the time required for the change in the angular velocity to occur.

Answers

Answer:

The time for the change in the angular velocity to occur is 14.08 secs

Explanation:

From the question,

the angular acceleration is - 4.46 rad/s²

Angular acceleration is given by the formula below

\alpha =(\omega -\omega _(o) )/(t - t_(o) )

Where \alpha is the angular acceleration

\omega is the final angular velocity

\omega _(o) is the initial angular velocity

t is the final time

t_(o) is the initial time

From the question

\alpha = - 4.46 rad/s²

\omega _(o) = 0 rad/s (starting from rest)

\omega = -31.4 rad/s

t_(o) = 0 s

Now, we will determine t

From \alpha =(\omega -\omega _(o) )/(t - t_(o) ), then

-4.46 = (-31.4 - 0)/(t - 0)

-4.46 = (-31.4)/(t)

t = (-31.4)/(-4.46)

t = 7.04 secs

This is the time spent in one direction,

Since the angular displacement of the wheel is zero ( it returned to its initial position), then the time required for the change in the angular velocity will be twice this time, that is 2t

Hence,

The time is 2×7.04 secs = 14.08 secs

This is the time for the change in the angular velocity to occur.

What is the weight of a 5.0-kilogram object at the surface of Earth?A
5.0 kg
B
25 N
C
49 kg
D
49 N

Answers

The weight of a 5.0-kilogram object at the surface of Earth is 49 N. Hence, option (C) is correct.

What are mass and weight?

Mass is a fundamental quantity in physics and the most fundamental attribute of matter. Mass can be defined as the amount of matter contained in a body. Kilogram is the SI unit of mass (kg).

A body's mass does not alter at any time. Only in severe instances where a massive amount of energy is provided or removed from the body.

The force of gravity acting on a body is measured by weight. The weight formula is as follows: w = mg. Because weight is a force, it has the same SI unit as force; the SI unit of weight is Newton (N).

The weight of the  object is = 5.0 × 9.8 N

= 49.0 N.

Hence, option (B) is correct.

Learn more about weight here:

brainly.com/question/10069252

#SPJ2

Answer:

49 N (d)

Explanation:

w= mg = 5 kg * 9.8 m/s^2 = 49 N

What displacement do I have if I travel at 10 m/s E for 10 s? A. 1 m E B. 1 m C. 100 m D. 100 m E Scalar quantities include what 2 things? A. Number and direction B. Numbers and units C. Units and directions D. Size and direction What measures distance in a car? A. Odometer B. Pressure gauge C. Speedometer D. Steering wheel What displacement do I have if I travel 10 m E, then 6 m W, then 12 m E? A. 28 m E B. 16 m E C. 16 m D. 28 m

Answers

Hope this will help you

Final answer:

The displacement is 100 m to the east.

Explanation:

The displacement can be calculated using the formula:

Displacement = Velocity × Time

In this case, the velocity is 10 m/s to the east and the time is 10 seconds.

So, Displacement = 10 m/s × 10 s = 100 m to the east.

Learn more about Displacement here:

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A car travels along a straight line at a constant speed of 53.0 mi/h for a distance d and then another distance d in the same direction at another constant speed. the average velocity for the entire trip is 26.5 mi/h. (a) what is the constant speed with which the car moved during the second distance d?

Answers

A distance of d is covered with 53 mile/hr initially.Time taken to cover this distance t1 = d/53 hourNext distance of d is covered with x mile hours.Time taken to cover this distance t2 = d/x hours.We have average speed = 26.5 mile / hour          

                                         = Total distance traveled/ total time taken                  

                                         = (2d)/((d)/(53)+(d)/(x)) = (2)/((1)/(53)+(1)/(x) )  = (106x)/(x+53)

                              26.5 = (106x)/(x+53) \n \n 79.5 x = 1404.5\n \n x = 17.67 miles/hour

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes the amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear

Answers

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × e^{(-bt)/(2m)}  × cos(ωt + ∅ )    ..................................... ( 1 )          

here function A × e^{(-bt)/(2m)}   is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .

Answers

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

x(t)=(0.28\ m)\cos[(8\ rad/s)t]....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

y=A\omega \cos\omega t

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.