True or False: If a point charge has electric field lines that point into it, the charge must be positive.

Answers

Answer 1

Answer:

False

Explanation:

If a point charge has electric field lines point into it,then charge must be negative because electric lines point into negative charges and point out of positive charges


Related Questions

You are driving your car at a speed of 19.0 m/s and you hit the brakes. The car accelerates at -3.50 m/s2. (a) How long does it take the car to cover 10.0 m? (b) What is the final velocity of the car?

Answers

Recall the formulas,

[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]

[tex]v_f=v_i+at[/tex]

(a)

[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]

(b)

[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]

[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]

A rod attracts a positively charged hanging ball. the rod is?a) negativeb) positivec) neutrald) either negative or neutrale) either positive or neutral

Answers

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.

Consider two balls in motion at the same time. Joe drops the first ball from rest at height h. Directly below Joe, on the ground, Hayley simultaneously tosses a second ball upward with speed v0.a. If the two balls collide at the moment the second ball is instantaneously at rest, what is the height of the collision?b. What is the relative speed of the balls when they collide?

Answers

Answer:

A. To find height of collision let's find

Speed of collision first

Using

Vf= √2g(h-g)

But for Harvey's ball we have

V²-2gy= 0

y=( v²/2g) so this is height of collision

B.

To find relative speed of the ball

Using V at y to find Vsr

Vs= √2g(h-v²/2g)

= √2gh- v²

2 differences between calorimeter and thermometer ?

Answers

Answer:

Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.

Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).

Explanation:

20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream shop in 10 minutes. You spend 15 minutes ordering your ice cream and then return home. Since you have an ice cream cone in your hand, it takes 20 minutes to walk home. 1. Find the total displacement and total distance traveled. 2. Find your average speed and average velocity in meters/min. 3. Find your average speed and average velocity in meters/sec.

Answers

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

I point
Is the car's speed increasing or decreasing with time?
not enough information
decrease
increase
constant

Answers

Answer:

It's increasing with time

It’s increasing b/c the dots are moving further

chemical reaction to make a foam?​

Answers

Answer: Hydrogen Peroxide

Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.

A car originally traveling at 30.0 m/s manages to brake for 5.0 seconds while traveling 125 m along a road. After those first 5.0 seconds, the brakes fail. After an additional 5.0 seconds it travels an additional 150 m further down the road. What was the magnitude of the acceleration of the car after the brakes failed

Answers

Answer:

The magnitude of the acceleration of the car after the brakes failed is 4 m/s²

Explanation:

The car was originally traveling at 30.0 m/s, that is

The initial velocity, [tex]u[/tex] = 30.0 m/s

The time spent while the car manages to brake is 5.0 seconds, that is

time, [tex]t[/tex] = 5.0 secs

and the distance traveled during this time is

distance, [tex]s[/tex] = 125 m

From one of the equations of kinematics for linear motion,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

Where [tex]a[/tex] is the acceleration

We can determine the deceleration of the car during the first 5.0 seconds

Hence,

From,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

[tex]125 = 30.0(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]125 =150.0 + 12.5a[/tex]

[tex]12.5a = 125 - 150.0[/tex]

[tex]12.5a = -25\\a = \frac{-25}{12.5}\[/tex]

[tex]a = - 2.0 m/s^{2}[/tex] (Negative sign indicates deceleration)

Now we will calculate the final velocity reached at this time

From,

[tex]v^{2} = u^{2} + 2as[/tex]

Where [tex]v[/tex] is the final velocity

[tex]v^{2} = 30.0^{2} + 2(-2.0)(125)\\v^{2} = 400\\v = \sqrt{400} \\v = 20 m/s \\[/tex]

This is the final velocity reached by the car during the first 5.0 seconds

Now, for the magnitude of the acceleration of the car after the brakes failed,

After the brakes failed,

it travels an additional 150 m further down the road, that is

s = 150m

an additional 5.0 seconds, that is

t = 5.0 seconds

Also, from

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex]

The initial velocity here will be the final velocity for the first 5.0 seconds, that is,

u = 20 m/s

Hence,

[tex]s = ut + \frac{1}{2}at^{2} \\[/tex] becomes

[tex]150 = 20(5.0) + \frac{1}{2}(a)(5.0)^{2}[/tex]

[tex]150 = 100 + 12.5a\\12.5a = 150 - 100\\12.5a = 50\\a = \frac{50}{12.5} \\a = 4m/s^{2}[/tex]

Hence, the magnitude of the acceleration of the car after the brakes failed is 4 m/s²

How long did it take me to make this?

Answers

1hour because that takes a lot of time

Answer:

Its according from what it is made with if wax it takes longer but if with a craft paper it takes lesser time if paper approximately 20 to 25 minutes

Find the magnitude of the magnetic field ∣∣B⃗ (r)∣∣ inside the cylindrical resistor, where r is the distance from the axis of the cylinder, in terms of i, r, r0, l, and other given variables. You will also need π and μ0. Ignore fringing effects at the ends of the cylinder.

Answers

Answer:

The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

Explanation:

Given that,

Distance from the axis of the cylinder = r

We need to calculate the magnetic field inside the cylindrical resistor

Using formula of magnetic field

[tex]\oint{\vec{B}\cdot\vec{dl}}=\mu_{0}i_{encl}[/tex]

[tex]B\cdot(2\pi r)=\mu_{0}\dfrac{i\pir^2}{\pi r_{0}^2}[/tex]

Where, r₀ = radius

r = distance

i = current

[tex]|\vec{B}(r)|=\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

Hence, The magnetic field inside the cylindrical resistor is [tex]\dfrac{\mu_{0}ir}{2\pir_{0}^2}[/tex]

A typical atom has a diameter of about 1.0×10−10m.
Approximately how many atoms are there along a 2.0 −cm line?
Express your answer using two significant figures.

Answers

Answer:

10m

Explanation:

if it was at the 2.0 line it would be 10 m

If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then there are approximately atoms are there along a 2.0-centimeter line.

What are significant figures?

In positional notation, significant figures refer to the digits in a number that is trustworthy and required to denote the amount of something, also known as the significant digits, precision, or resolution.

As given in the problem If a typical atom has a diameter of about 1.0×10⁻¹⁰ m, then we have to find out approximately how many atoms are there along a 2.0-centimeter line,

diameter of the one atom =  1.0×10⁻¹⁰

approximate number of atoms in 2 cm line = 2 ×10⁻² /( 1.0×10⁻¹⁰ )

                                                                         =2 ×10⁸ atoms

Thus, there are approximately 2 ×10⁸ atoms are there along a 2.0-centimeter line.

Learn more about significant figures here,

brainly.com/question/14359464

#SPJ2

hows a map of Olivia's trip to a coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8 mi to where Broadway turns, and then continues another 1.4mi to the shop.

Answers

The question is incomplete. Here is the complete question.

The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.

What is the magnitude of the total displacement of her trip?

Whta is the direction of the total displacement of her trip?

Answer: Magnitude = 2.6mi

              Direction: 54.65° east

Explanation: Displacement is the change in postition of a moving object.

There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.

For this method, total displacement is given by:

[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]

[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]

[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;

[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;

θ is the angle the resulting displacement;

For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:

[tex]sin30=\frac{x}{1.4}[/tex]

[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]

x = 0.7

Then,

[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7

[tex]\Delta d_{x}[/tex] = 1.5

Related to y, there are no y-component in the second part of Olivia's trip and for the third part:

[tex]cos30=\frac{y}{1.4}[/tex]

[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]

y = 1.21

Then,

[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21

[tex]\Delta d_{y}[/tex] = 2.11

Total displacement is

[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]

[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]

[tex]\Delta d_{t}=[/tex] 2.6

Magnitude of Olivia's total displacement is 2.6mi

On the map, joining the initial and final points gives a vector pointing towards east at angle:

[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]

[tex]\theta=tan^{-1}(1.41)[/tex]

θ = 54.65°

Direction of total displacement is 54.65° East.

Answer:

1.9 mi, 350.7°

Explanation:

If continued another 1.2 mi*

solving for x

0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84

solving for y

-0.9 + 0 + 1.2(sin30) = -0.3

^negative because it is going downward

a) solving for magnitude

[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]

b) solving for direction of total displacement

[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]

∘, measured counterclockwise from the eastward direction

360 - 0.16 = 359.84°

*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi

From the gravitational law, calculate the weight W (gravitational force with respect to the earth) of a 70 kg spacecraft traveling in a circular orbit 275 km above the earth's surface. Express W in Newtons and pounds.

Answers

Answer:

The  value in Newton is [tex]W =  631.92 \  N[/tex]

The  value in pounds is    [tex]W  = 142 \ lb[/tex]

Explanation:

From the question we are told that

  The  mass of the spacecraft is  [tex]m =  70 \  kg[/tex]

   The distance above  the earth is  [tex]d =  275 \  km  =  275000 \  m[/tex]

Generally the gravitational force with respect to the earth is mathematically represented as

       [tex]W =  \frac{G * m *  m_e}{ (d + r_e)^2}[/tex]

Here [tex]m_e[/tex] is the mass of earth with value [tex]m_e =  5.978 *10^{24} \  kg[/tex]

       [tex]r_e[/tex] is the radius of the earth with value  [tex]r_e  =  6371  \ km  =  6371000 \ m[/tex]

   G is the gravitational constant with value [tex]G  =  6.67 *10^{-11}  \  m^3/ kg\cdot s^2[/tex]

So

     [tex]W =  \frac{ 6.67 *10^{-11} *  70 *  5.978 *10^{24}}{ (275000 + 6371000)^2}[/tex]

     [tex]W =  631.92 \  N[/tex]

Converting to  pounds

    [tex]W =  \frac{631.92  }{4.45}[/tex]

        [tex]W  = 142 \ lb[/tex]

     

Peter left Town A at 13:30 and travelled towards Town B at an
average speed of 40 mph. At 13:45, Philip left Town A for Town
B at an average speed of 30 mph. What was the distance
between them at 15:00?

Answers

Answer:

Explanation: From 13:30 to 15:00, it past: 1 h 30 mins = 1.5

Then, the distance covered by Peter: 40x1.5= 60 miles

From 13:45 to 15:00, it pasts; 1 h 15min =1.25

Then, the distance covered by Philip. 30 x 1.25 = 37.5 miles

Lastly, the distance between them: 60-37.5= 22.5 miles

So the answer is 22.5

Write all the different ways you can think of that describe what it means to be healthy

Answers

Answer:

Eat more healthy foods.  Workout and build your immune system.

Explanation:

Eat  Healthy foods like Carrots, Apples, Bannas, Pears, and anything that deals with not much of any sugar. An example of unhealthy foods is Cakes, Chocolates, Candy, and more. Drink a lot of water.

Common benefits of lower body endurance include improved

Answers

Answer:

Hearing, Vision and Metabolism.

A bird flutters around in a tree in a path described by the dark line: ​ ​ Which vector represents the average velocity of the bird? Question 2 options:

Answers

Answer:

Displacement vector represents the average velocity of the bird.

Explanation:

Given that,

A bird flutters around in a tree in a path described by the dark line.

Suppose, given vectors

(a). Time, (b). Displacement, (c). speed, (d). distance

We know that,

Vector quantity :

Vector quantity has direction and magnitude.

Average velocity :

Average velocity is equal to the displacement divided by time.

In mathematically,

[tex]v=\dfrac{D}{t}[/tex]

Where, D = displacement

t = time

We need to find the vector which is represents the average velocity of the bird

Using given data

Average velocity of the bird shows the displacement over the time.

Displacement is the vector quantity.

Hence, Displacement vector represents the average velocity of the bird.

A ship sets out to sail to a point 116 km due north. An unexpected storm blows the ship to a point 121 km due east of its starting point. (a) How far (in km) and (b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination?

Answers

Explanation:

In this problem, we are meant to slove for the resultant and the direction of the the vectors given

Given data

let the sail to a point due north be y= 116km

and the point due east be x= 121 km

(a) How far (in km)

The resultant between the two points is the distance between them

[tex]r=\sqrt{x^2+y^2} \\\\r=\sqrt{121^2+116^2} \\\\r=\sqrt{14641+13456} \\\\r=\sqrt{28097} \\\\r=167.62[/tex]

The distance between the points is 167.62

(b) in what direction (as an angle from due east, where north of east is a positive angle) must it now sail to reach its original destination

the direction can be gotten using

tan∅= y/x

∅= tan-1 (y/x)

∅= tan-1(116/121)

∅= tan-1(0.958)

∅= 43.77°

The direction is 43.77°

The main force(s) acting on the puck after receiving the kick is (are):_________.A) a downward force of gravity and an upward force exerted by the surfaceB) a downward force of gravity, and a horizontal force in the direction of motionC) a downward force of gravity, an upward force exerted by the surface, and a horizontal force in the direction of motionD) a downward force of gravityA) a downward force of gravity and an upward force exerted by the surface

Answers

Answer:

the statements, the correct one is A

a downward force of gravity and an upward force exerted by the surface

Explanation:

When the disc is hit, a thrust force is exerted in the direction of movement, at the moment the disc moves this force loses contact and becomes zero.

When the movement is already established there are two main forces: gravity that acts downwards and the reaction force to the support of the disk called normal that acts upwards.

As it is not mentioned that there is friction, this force that opposes the movement is zero.

Analyzing the statements, the correct one is A

What are the benefits of testing your heart rate during exercise

Answers

Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.

Explanation:

Answer:

You're getting exercise and training your heart!

Explanation:

Which of the following is not a valid use of your driver's license?
O proof of your ability to operate a motor vehicle
O proof of your age
proof of your residency
O proof that you have liability insurance
NEXT QUESTION

Answers

In the state where I live, your driver's license is not a proof that you have liability insurance.  You don't need liability insurance to get a driver's license, but you need it in order to operate a car that you own.

It may be different in the state where YOU live.

NEXT QUESTION

A car going initially with a velocity 13.5 m/s accelerates at a rate of 1.9 m/s 2 for 6.2 s. It then accelerates at a rate of -1.2 m/s2 until it stops. a) Find the car’s maximum speed b) Find the total time from the start of the first acceleration until the car is stopped c) What’s the total distance the car travels?

Answers

(a) For the first 6.2 s, the car has velocity at time t given by

[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after 6.2 s, it attains a velocity of

[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]

For any time t after 6.2 s, its velocity is given by

[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.

(b) Solve for t (after 6.2 s) that makes v(t) = 0 :

[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]

[tex]\implies t\approx21.067\,\mathrm s[/tex]

It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.

(c) For the first 6.2 s, the car undergoes a displacement at time t of

[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]

[tex]\implies x\approx120.218\,\mathrm m[/tex]

For time t beyond 6.2 s, its displacement is

[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is

[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]

so the car travels a total distance of about 387 m.

A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 232 km and a direction 30.0o north of east. The displacement vector B for the second segment has a magnitude of 168 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle θ with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle θ.

Answers

Answer:

a) 121 km

b) 74°

Explanation:

To start with, we assume that there exist two components, the East and the North. We'd be representing the East, by "e" and the North, by "n"

Now we start with the

First vector:

east1 = 232 cos 30 = 201

north1 = 232 sin 30 = 116

Now, that of the second vector will be

east 2 = - 168

north2 = 0

Next, we add the two together and get

East components

201 - 168 = 33 east

North components

116 + 0 = 116

Therefore, the magnitude has to be

magnitude = √(33² + 116²)

Magnitude = √14545

Magnitude = 121

tanθ = 116/33

Tanθ = 3.51

θ = tan^-1 3.51

θ = 74° North East

What line on a wheather map indicates áreas where the temperature is the same?

Answers

Answer:Isobars and isotherms

Explanation:

What is the value of the radius of the following circle with an area of 154 cm2?

Answers

The area of ANY circle is (π) · (radius²).

So ...

(π) · (radius²) = 154 cm²

radius² = (154 cm²) / (π)

radius² = 49.02 cm²

radius = √(49.02 cm²)

radius = 7 cm

Answer:

[tex] \boxed{\sf Radius \ of \ circle \ (r) = 7 \ cm} [/tex]

Given:

Area of circle = 154 cm²

To Find:

Radius of circle (r)

Explanation:

[tex] \bold{Area \: of \: circle = \pi r^2}[/tex]

[tex] \sf \implies \pi {r}^{2} = 154 \\ \\ \sf \implies {r}^{2} = \frac{154}{\pi} \\ \\ \sf \implies {r}^{2} = \frac{154}{ \frac{22}{7} } \\ \\ \sf \implies {r}^{2} = \frac{154 \times 7}{22} \\ \\ \sf \implies {r}^{2} = \frac{1078}{22} \\ \\ \sf \implies {r}^{2} = 49 \\ \\ \sf \implies {r}^{2} = {7}^{2} \\ \\ \sf \implies r = \sqrt{ {7}^{2} } \\ \\ \sf \implies r = 7 \: cm[/tex]

You are standing on top of the Empire State Building (height 330 [m]) when Superman flies past. He is headed straight down with a steady speed of 35 [m/s]. (Superman can do things like this.) At the instant he goes past, you drop a 10 [kg] lead ball over the edge. At what height above the sidewalk does the ball pass Superman?

Answers

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

[tex]S=ut[/tex]...(I)

For ball,

[tex]S=u't+\dfrac{1}{2}gt^2[/tex]

From equation (I) and (II)

[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]

[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]

[tex]u=\dfrac{1}{2}\times g\times t[/tex]

[tex]t=\dfrac{2u}{g}[/tex]

Put the value into the formula

[tex]t=\dfrac{2\times35}{9.8}[/tex]

[tex]t=7.1\ sec[/tex]

We need to calculate the distance travelled

Using formula of distance

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=35\times 7.1[/tex]

[tex]d=248.5\ m[/tex]

Hence, The ball will be overtake superman after travelled distance 248.5 m.

A sidewalk has a length of 75.00m. How many inches is this? (Hint: you need to use two unit conversion fraction. 1 cm equals about 0.3937 inches)

Answers

Length = (75.00 m)

Length = (75 meter) x (3.28084 foot/meter) x (12 inch/foot)

Length = (75 x 3.28084 x 12) (meter-foot-inch / meter-foot)

Length = 2,952.76 inches

Check Concepts
4.
35. Which of the following do you calculate
when
you
divide the total distance trav-
eled by the total travel time?
A) average speed
B) constant speed
C) variable speed
D) instantaneous speed

Answers

Answer:

I think its A.........

Two equal charges are 2m2m apart. If the charges and the distance are divided by two, how is the force between the charges affected? g

Answers

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]

Therefore, the force between the charges are not affected.

Is the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth?

Answers

Answer:

Yes

Explanation:

This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic

Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

Gravitational force is the force by which earth attracts other objects by mass.The electrostatic force is the force of an object due to charge.Electrostatic forces are much stronger than gravitational forces. because gravity depends on mass, atoms have less masses so that the gravitational forces between them is close to zero. Whereas, the electrostatic force related to charges is bigger.

Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

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