Two 3 m long conductors are separated from each other by 5 mm and carry a current of 10 A dc. Calculate the force that one conductor exerts on the other.

Answers

Answer 1

Answer:

0.012 N

Explanation:

The formula to apply is that adopted from the Ampere law which is;

F= μ* I₁*I₂*l /2πd   where

F is force that one conductor exerts on the other.

μ = magnetic permeability of free space = 4π×10⁻⁷ T. m/A

I₁ = current in conductor one=10 A

I₂ = current in conductor two= 10 A

l= length of conductor= 3 m

d= distance between the conductors = 5 mm = 0.005 m

Applying the values in the equation

F= 4π×10⁻⁷ *10*10*3 / 2π*0.005

F= 6 * 10⁻⁵ / 0.005

F=0.012 N


Related Questions

A dwari planet is a:

small rocky object that orbits the Sun and is usualy found in a bet beiween me orbits of Mars and Jupiter.

small object made of ice and dust thai orbits the Sun and forms a coma as i approaches the Sun.

round object that orbits the Sun but lacks the abity to clear the neghborhood around is orbit.

meteoroid thai bums up in Earth's amosphere, producing a streak of light.​

Answers

Answer:

round object that orbits the Sun but lacks the ability to clear the neighborhood around is orbit.

Explanation:

It is true that a dwarf planet is a round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit.

Due to the size of a dwarf planet, it does not possess enough gravitational attraction or force to clear the orbit around it or other bodies.

One of the notable dwarf planets is Pluto. It was formerly thought to be planet but it has since been delisted.

Pluto's orbit lies beyond that of Neptune.

You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of force you applied to your friend.

Answers

Answer:

4

Units:

Newtons

The velocity of a Froghopper flea, which is represented by a vector, has a direction and a magnitude. If a coordinate system is oriented where the x-axis is horizontal, and the y-axis is vertical, is it possible to write expressions for the x and y components of the velocity vector in terms of the vector's magnitude and direction.

Answers

Answer:

Explanation:

The solution of the question cab e found in attachment below:

a car traveled at 100 km h for 3 hours how far did it travel

100 km

300 km/h

300 km

3 km/h

Answers

Answer:

300km

Explanation:

Given parameters:

Speed of the car  = 100km/h

Time taken for the travel  = 3hrs

Unknown:

How far did it travel  = ?

Solution:

To solve this problem, we must understand that;

        Speed  = [tex]\frac{distance}{time}[/tex]  

 Distance  = speed x time

 Distance  = 100km/hr x 3hr = 300km

Exercise 3 In an equation y mx+c; y and x have dimensions of length and c is constant. What are the dimensions of m'
A. Mass
B. Length
C. Time
D. m is Dimensionless

Answers

Answer:

D. m is Dimensionless

Explanation:

The equation of a straight line is given as:

        y  = mx + c

Dimension of y = l

                       x = l

                       c has no dimension

 So;

if we do a dimensional analysis:

            L  = m L + 0

              m  = 1

So, m has no dimension

A child is twirling a 0.0113-kg ball on a string in a horizontal circle whose radius is 0.147 m. The ball travels once around the circle in 0.388 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase

Answers

Answer:

0.435 N

The centripetal force increases by a factor of 4

Explanation:

Let the centripetal force be F

F =mv^2/r

m= mass of the object

v = linear velocity of the object

r = radius of the circular path

but v = 2πr/T = 2 * 3.142 * 0.147 / 0.388

v = 2.38 m/s

F = 0.0113 * (2.38)^2/0.147

F= 0.435 N

if v= 2v

Then;

F = m(2v)^2/r

F = m4v^2/r

F= 4mv^2/r

The centripetal force increases by a factor of 4

A hockey puck slides across the ice and eventually comes to a stop. Why did the puck stop?

Answers

Answer: the total energy of the puck, ice surface, and surrounding air decreases to zero

Explanation:

A hockey puck slides across the ice and eventually comes to a stop because of friction between surface of the puck and ice surface.

What is Friction ?

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force. frictional force is directly proportional to the Normal(N). i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Frictional force is equal to

[tex]F_{fri}[/tex] = μmg

When hockey puck slides across the ice, friction between surface of puck and surface of ice produces resistance to the motion of the puck, due to resistance puck slow down slowly and eventually come to a stop. Motion and frictional force are opposite to each other. we can calculated the exact value of frictional force when we know the coefficient of friction between puck and ice surface.

hence due to friction, puck come to a stop

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A baby crawls 3m in 6s. The baby travels toward the west. Calculate the velocity of the baby.

Answers

Answer:

.5m per second to the west

Explanation:

distance divided by time

5m per second to the west ooo

You are arranging family members for a photograph, and you want them to stand in order of increasing height. Your uncle is half a foot shorter than your aunt, who is taller than your cousin. Your grandmother is 2 in. shorter than your grandfather. Your brother is 1 in. taller than your aunt and 3 in. taller than your cousin. Your grandfather is 1 cm taller than your aunt.
List the relatives in order of increasing height.

Answers

Answer:

Uncle , Cousin, Grandmother , Aunt , Brother , Grandfather

Explanation:

Arranging in the order of increasing height simply means starting from the shortest to the tallest ( ascending order )

Number of relatives = 6

Uncle , Aunt , cousin, grandmother , grandfather, Brother

First lets write out the relationship between your relatives using a feet as our S.I unit

Uncle is shorter than Aunt by ;  1/2 feet = 0.5 feet

Cousin is shorter  than Aunt by : unknown value

Cousin is shorter than Brother by ; 1/4 feet = 0.25 feet

Grandmother is shorter than Grandfather by ; 0.16 feet

Aunt is shorter than Grandfather by ; 0.03 feet

Aunt is shorter than brother by ; 0.08 feet

Hence the list of relatives in ascending order

Uncle , Cousin, Grandmother , Aunt , Brother , Grandfather

(a)Uncle is taller than cousin.

(b)Relatives in order of increasing height will be Uncle , Cousin, Grandmother , Aunt , Brother , Grandfather.

What is ascending order ?

Ascending order is a way of organizing integers in ascending order from lowest to biggest value. The sequence is from left to right.

The given number of relative is 6 which is uncle , aunt , cousin, grandmother , grandfather, brother.

1.Uncle is shorter than Aunt by ;  1/2 feet = 0.5 feet

2.Cousin is shorter  than Aunt by : unknown value

3.Cousin is shorter than Brother by ; 1/4 feet = 0.25 feet

4.Grandmother is shorter than Grandfather by ; 0.16 feet

5.Aunt is shorter than Grandfather by ; 0.03 feet

6.Aunt is shorter than brother by ; 0.08 feet

Hence relatives in order of increasing height will be Uncle , Cousin, Grandmother , Aunt , Brother , Grandfather.

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A student is transmitting sound waves through various materials. Through which metal in the table will the sound waves travel the fastest? Aluminum Copper Lead Brass (70% Cu, 30% Zn) 6,420 4,700 5,010 1,960 Speed of sound (m/s) O
A. Brass O
B. Aluminum
C. Copper O
D. Lead​

Answers

Answer:

Aluminum

Explanation:

promise

give  me brainlest plleaseee

Answer:aluminum

Explanation:

The volume of a solid object tends to increase as its temperature increases. Which quantities determine how large the change in volume is

Answers

Answer:

I. The change in temperature of the object.

II. The initial volume of the object.

Explanation:

Generally, the volume of a solid object tends to increase as its temperature increases and this phenomenon is known as thermal expansion.

Hence, the quantities which determine how large the change in volume of a solid object is includes;

I. The change in temperature of the object.

II. The initial volume of the object.

This ultimately implies that, when a solid object is heated, the atoms of the object vibrate rapidly about their fixed points and thus, causing an increase in the volume of the object.

In conclusion, this scientific phenomenon known as thermal expansion is valid and true for all the three (3) states of matter;

Solid. Liquid. Gas.

2. A girl of weight 400 N has feet of area 100 cm2. What

pressure does she put on the ground ?

Answers

Answer:

Explanation:

Given:

Weight = 400N

Area = 100cm^2

If 1cm^2 = 1×10^-4m^2

Hence 100 cm^2 = 0.01m^2

Pressure = force/area .........1

Force = ma = mg which is also the weight of the girl. Hence substitute the values into 1

P = 400/0.01

P = 40000 N/m^2

The pressure she puts on the ground is 40000 N/m^2

In an experiment, a cement block of mass 14 kg was placed on top of 8 (carefully arranged) eggs without breaking the eggs.
1) Calculate the force per egg caused by the weight of the block.
2) If the pressure on each egg is 25 N/mm 2 , calculate the contact area, in mm 2 , between the block and one egg.

Answers

Answer:

1) The force per egg caused by the weight of the block is 17.15 N

2) [tex]A= 0.686\ mm^2[/tex]

Explanation:

Pressure

The pressure is computed as the force acting upon a surface divided by the area of the surface :

[tex]P=\frac{F}{A}[/tex]

The weight of an object of mass m is:

W = mg

Where g is the gravitational acceleration of [tex]9.8\ m/s^2[/tex]

The cement block has a mass of m=14 kg, thus its weight is:

W = 14*9.8

W = 137.2 N

1) This weight is evenly distributed over the 8 eggs, thus the force per egg is:

F = 137.2 N/8= 17.15 N

The force per egg caused by the weight of the block is 17.15 N

2) It's given the pressure on each egg as [tex]P= 25 \ N/mm^2[/tex] and we know the force acting on each one is F=17.15 N. To find the contact area we solve the formula of the pressure for A:

[tex]\displaystyle A=\frac{F}{P}[/tex]

[tex]\displaystyle A=\frac{17.15}{25}[/tex]

[tex]\mathbf{A= 0.686\ mm^2}[/tex]

Ball X of mass 1.0 kg and ball Y of mass 0.5kg travel toward each other on a horizontal surface. Both balls travel with a constant speed of 5 m/s until they collide. During the collision, ball Y exerts an average force with a magnitude of 40N for 1/6s on ball X . Which of the following best predicts ball momentum after the collision?
A. Ball Y will travel at a speed less than 5 m/s in the same direction of travel as before the collision.
B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.
C. Ball will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.
D. Ball Y motion cannot be predicted because the impulse on it is not known.

Answers

Answer:

C

Explanation:

Ball Y will travel at a speed greater than 5 m/s in the opposite direction of travel as before the collision.

The ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

The given parameters;

mass of ball X = 1 kgmass of ball Y = 0.5 kgspeed of the two balls after collision, v = 5 m/sforce exerted by ball Y, = 40 Ntime, t = 1/6 s

Apply the principle of conservation of linear momentum. The total momentum of the balls before collision must equal total momentum after collision.

[tex]m_xv_x_1 + m_yv_y_1 = m_x v_x_2 + m_yv_y_2\\\\m_xv_x_2 - m_xv_x_1 = -(m_yv_y_2 - m_yv_y_1)\\\\\Delta P_x = - \Delta P_y[/tex]

The impulse experienced by ball Y is calculated as follows;

[tex]\Delta P_y = m_yv_y_2- m_yv_y_1 = Ft = 40 \times \frac{1}{6} = 6.67 \ kgm/s[/tex]

The final speed of ball Y is calculated as follows;

[tex]m_yv_y_2 - m_yv_y_1 = 6.67 \\\\m_yv_y_2 = 6.67 + m_yv_y_1\\\\0.5(v_y_2) = 6.67 + (0.5 \times 5)\\\\0.5(v_y_2) = 9.17\\\\v_y_2 = \frac{9.17}{0.5} \\\\v_y_2 = 18.34 \ m/s[/tex]

Thus, the ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

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A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?

(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)

Answers

i believe the answer is (3)

The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.

The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.

From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.

where;

|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile

Using trigonometric approach for ΔOAP;

[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]

[tex]\mathbf{AP =OP\times sin \theta}}[/tex]

AP = 200 × sin 30°

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A pendulum with a mass of 1 kg is released from a height of 1.5 cm above the height of its resting position. How fast will the pendulum be moving when it passes through the lowest point of its swing?

Answers

Answer:

v = 0.54 m/s

Explanation:

The Principle Of Conservation Of Mechanical Energy

In the absence of friction, the total mechanical energy is conserved. That means that

Em=U+K is constant, being U the potential energy and K the kinetic energy

U=mgh

[tex]\displaystyle K=\frac{mv^2}{2}[/tex]

When the mass (m=1 kg) of the pendulum is at the top of the path at a height of h=1.5 cm=0.015 m, its kinetic energy is 0 and its potential energy is:

U=1 * 9.8 * 0.015 = 0.147 J

That potential energy is completely transformed into kinetic energy at the bottom of the swing. The speed can be calculated by solving for v:

[tex]\displaystyle \frac{mv^2}{2}=0.147[/tex]

Multiplying by 2 and dividing by m:

[tex]\displaystyle v^2=\frac{2K}{m}[/tex]

[tex]\displaystyle v^2=\frac{2*0.147}{1}=0.294[/tex]

[tex]v=\sqrt{0.294}[/tex]

v = 0.54 m/s

5.How do simple machines make work easier?(1 point)
a.They decrease the distance needed between a person and an object, allowing the person to apply more force and do more work.

b.They magnify the force being applied, thereby increasing the work they can do with no extra energy exerted by the person using the machine.

c.They reduce the amount of force needed to move an object to do the same amount of work.

d.They increase the distance over which a force is needed, doing more work with less force.

Answers

The correct answer is d

Answer:

They increase the distance over which a force is needed, doing more work with less force.

Explanation:

A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.

Answers

Answer:

-22.1

Explanation:

1 / 4

Step 1. List horizontal (xxx) and vertical (yyy) variables

xxx-direction yyy-direction

t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text

a_x=0a  

x

​  

=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a  

y

​  

=−9.8  

s  

2

 

m

​  

a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction

\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text

v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v  

y

​  

=?v, start subscript, y, end subscript, equals, question mark

v_{0x}=120\,\dfrac{\text m}{\text s}v  

0x

​  

=120  

s

m

​  

v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v  

0y

​  

=0v, start subscript, 0, y, end subscript, equals, 0

Note that there is no horizontal acceleration, so v_x=v_{0x}v  

x

​  

=v  

0x

​  

v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.

Also, the package has no initial vertical velocity.

Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:

\Delta x=v_xtΔx=v  

x

​  

tdelta, x, equals, v, start subscript, x, end subscript, t

Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv  

y

​  

v, start subscript, y, end subscript:

\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared

Hint #22 / 4

Step 2. Find ttt from horizontal variables

\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}  

Δx

t

t

​  

 

=v  

x

​  

t

=  

v  

0x

​  

 

Δx

​  

 

=  

120  

s

m

​  

 

255m

​  

 

=2.125s

​  

 

Hint #33 / 4

Step 3. Find \Delta yΔydelta, y using ttt

Using ttt to solve for \Delta yΔydelta, y gives:

\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}  

Δy

​  

 

=v  

0y

​  

t+  

2

1

​  

a  

y

​  

t  

2

 

=  

(0)t

​  

+  

2

1

​  

(−9.8  

s  

2

 

m

​  

)(2.125s)  

2

 

=−22.1m

​  

 

Hint #44 / 4

The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.

A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?

Answers

Answer:

A Impulse = – 25 Ns

B. Time = 5 s

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

Time (t) =?

A. Determination of the Impulse.

Mass (M) = 5 Kg

Initial velocity (u) = 8 m/s

Final velocity (v) = 3 m/s

Impulse (I) =?

I = Ft = M(v – u)

I = M(v – u)

I = 5 (3 – 8)

I = 5 × – 5

I = – 25 Ns

NOTE: the negative sign indicates that the net force is acting in the negative direction.

B. Determination of the time.

Impulse (I) = 25 Ns

Force (F) = 5 N

Time (t) =?

I = Ft

25 = 5 × t

Divide both side by 5

t = 25 / 5

t = 5 s

Thus, it will take 5 s for the box to slide through the 15 m long ramp.

A dog sees a flowerpot sail up and then back past a window H high. If the total time the pot is in sight is t seconds, find the height above the window that the pot rises. (Let H = 2 m and t = 1.0 s, find the height above the window-- after you have found an algebraic solution.)

Answers

Answer:

maximum height = 0.1225 m

Explanation:

given data

H = 2m

t = 1 sec

solution

we consider here velocity of pot at lower side is u

and final velocity is v with acceleration a

time take is t/2

so

height h = u × [tex]\frac{t}{2}[/tex]  - 0.5 × g × [tex](\frac{t}{2})^2[/tex]     .................1

here

u = [tex]\frac{2}{t} \times ( h + \frac{gt^2}{8} )[/tex]

and

v = u +a t/2      .........................2

v = u + g t/2

v = [tex]\frac{2h}{t} + \frac{gt}{4} - \frac{gt}{2}[/tex]

v = [tex]\frac{2h}{t} - \frac{gt}{4}[/tex]

so that

maximum height is  = [tex]\frac{v^2}{2g}[/tex]

maximum height = [tex]\frac{(\frac{2h}{t} - \frac{gt}{4})^2}{2g}[/tex]

put here value of h and t

maximum height =  [tex]\frac{(\frac{2(2)}{1} - \frac{g(1)}{4})^2}{2g}[/tex]

maximum height = 0.1225 m

4. Friction is required for :
(a) Fast movement
(0) Both of the above
(b) Stopping the objec
(d) None of the above​

Answers

Both of the above! :)

Answer:

b) stopping the object

Explanation:

Friction always slows a moving object down. ... Friction can be a useful force because it prevents our shoes slipping on the pavement when we walk and stops car tyres skidding on the road. When you walk, friction is caused between the tread on shoes and the ground. This friction acts to grip the ground and prevent sliding

which of the following are not units used to measure energy?
a. joules
b. newtons
c. BTU
d. calories​

Answers

Answer:

The BTU, or British thermal unit, is actually a measure of heat.

When catching a baseball, a catcher's glove moves by 11 cm along the line of motion of the ball. If the baseball exerts a force of 464 N on the glove, how much work is done by the ball

Answers

Answer:

51.04 joules

Explanation:

The movement by the gloves = 11cm

Force by the baseball = 464N

First of all we are going to convert cm to metres

11cm = 11/100 meters = 0.11m

The formula for workdone is given as:

W = f x d

W = workdone

F = force

D = distance

Workdone = 464 x 0.11

= 51.04 joules

The workdone by the ball is 51.04 joules

Thank you

Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is

Answers

Answer:

The new force between them is increased by a factor of 16.

Explanation:

According to Coulombs law, the force of attraction between two (2) charges is given by the formula;

F = Kq1q2/r²

Given the following data;

q1 = 2q1

q2 = 2q2

r = r/2

Substituting into the equation, we have;

F = 2q1*2q2/(r/2)²

F = 4q1q2/r²/4

F = 4q1q2 * 4/r²

F = 16q1q2/r²

Therefore, the new force between them is increased by a factor of 16.

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 11 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?

Answers

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.We can find this distance simply applying the definition of average velocity, as follows:

       [tex]\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)[/tex]

The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       [tex]v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)[/tex]

where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².Solving for Δx, we get:

       [tex]\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)[/tex]

So, the total distance traveled was the sum of (1) and (3):Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.⇒Δx = Δx₁ + Δx₂ = 39 m. (5)Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       [tex]\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)[/tex]

Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        [tex]v_{omax} ^{2} = 2* a* \Delta x_{2} (7)[/tex]

Solving for Δx₂, we get:

       [tex]\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)[/tex]

Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.Taking the positive root in the quadratic formula, we get the following value for vomax:v₀max = 24.3 m/s.

Help can’t find the answer no where

Answers

i'm stuck on that question also

You have to deliver some 5.0-kg packages from your home to two locations. You drive for 2.0 h at 30 mi/h due east (call this segment 1 of your trip), then turn around and drive due west for 30 min at 28 mi/h (segment 2). Use a coordinate system with the positive x axis aimed toward the east and the origin at your home.
A. What is your position vector at the instant you reach the end of segment 1?
B. What is your position vector at the instant you reach the end of segment 2?
C. Calculate your displacement during segment 2.

Answers

Answer:

A. 60 mi.

B. 46 mi.

C. -14 mi.

Explanation:

A)

Assuming that we use a coordinate system with the positive x-axis aimed toward the east, and the origin at the start point, we can find the position vector at the end of the segment 1, applying the definition of average velocity, as follows:

       [tex]v_{avg1} = \frac{x_{1f} -x_{1o}}{t} (1)[/tex]

where x₁f = final position at the end of segment 1, x₁₀ = initial position at the start of the segment 1 = 0, t = time traveled during segment 1 = 2.0 h, and v₁, average velocity during segment 1 = 30 mi/h due east.Replacing by the givens, and solving for x₁f, we get:

        [tex]x_{1f} = v_{1avg} * t =30 mi/h * 2.0 h = 60 mi due east. (2)[/tex]

B)

In order to find the position at the end of the segment 2, we can use the same equation (1), but taking into account that the initial position will not be zero, but the final position at the end of the segment 1, i.e., 60 mi due east.Replacing by the givens, and solving for x₂f, we get:

       [tex]x_{2f} = x_{1f} + v_{2avg} * t = (-28 mi/h) * 0.5 h = 60 mi - 14 mi = 46 mi (3)[/tex]

C)  

The displacement during the segment two, is simply the difference between the final and initial positions for this segment.Since x₂₀= x₁f = 60 mi, and x₂f = 46 mi, we find that the displacement is as follows:Δx = x₂f - x₂₀ = 46 mi - 60 mi = - 14 mi.

Padded dashboards in cars are safer in an accident than nonpadded ones because an occupant hitting the dash has:________.
(a) increased time of impact.
(b) decreased impulse.
(c) decreased impact force.
(d) a and b above.
(e) a and c above.

Answers

Answer:

(e) a and c above.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = Mass * Velocity [/tex]

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

Padded dashboards in cars are safer in an accident than non-padded ones because an occupant hitting the dashboard of an automobile car has an increased time of impact and a decreased impact force because the force or shock experienced is high and happens rapidly over a short period of time, thus, the occupant has less time and velocity while absorbing the momentum of the car in the course of the collision.

Two charged particles attract each other with a force of magnitude F. If the distance between the charges is made 3.5 times as large and the charge on one of the particles is made 4 times as big, what is the ratio of the new F to the old one

Answers

Answer:

The ratio between the forces is:

[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]

Explanation:

The electrostatic force equation is:

[tex]F_{old}=k\frac{q_{1}q_{2}}{d^{2}}[/tex]

Where:

q1 and q2 are the electric charges

d is the distance between them

k is the electrostatic constant

Now, the distance is 3.5 times as large and q1 is 4 times as big, then the new force will be:

[tex]F_{new}=k\frac{4q_{1}q_{2}}{(3.5d)^{2}}[/tex]

[tex]F_{new}=\frac{4}{3.5^{2}}k\frac{q_{1}q_{2}}{d^{2}}[/tex]

We can rewrite this equation in terms of F(old)

[tex]F_{new}=\frac{4}{3.5^{2}}F_{old}[/tex]

Therefore, the ratio between the forces is:

[tex]\frac{F_{new}}{F_{old}}=\frac{4}{3.5^{2}}[/tex]

[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]

I hope it helps you!

In which three ways does the pattern seen on the screen during a double-slit
experiment support the wave model of light?
O A. All of the light that enters the slits passes through them.
B. The light bends around the edges of the slit.
C. There are bands of dimmer light.
D. There are bands of brighter light.

Answers

Answer:

b,c,d

Explanation:

the light bends around the edges of the slit

there are bands of brigter light

there are bands of dimmer light

( just took the quiz )

Option B,C and D shows the three ways does the pattern seen on the screen during a double-slit experiment.

What is double-slit experiment?

The double-slit experiment shows that light and matter may have properties that are both conventionally defined waves and particles, as well as the essentially probabilistic nature of quantum mechanical processes.

The three ways does the pattern seen on the screen during a double-slit experiment support the wave model of light;

B. The light bends around the edges of the slit.

C. There are bands of dimmer light.

D. There are bands of brighter light.

Hence, B,C and D are the three ways does the pattern seen on the screen during a double-slit experiment.

To learn more about the double-slit experiment, refer to the link;

https://brainly.com/question/17167388

#SPJ2

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