A car sits in an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 19 m/s when it reaches the end of the 110-m-long ramp. The traffic on the freeway is moving at a constant speed of 19 m/s. What distance does the traffic travel while the car is moving the length of the ramp

Answer:

Explanation:

Considering non - relativistic approach : ----

Speed of electron = 1 % of speed of light

= .01 x 3 x 10⁸ m /s

= 3 x 10⁶ m /s

Kinetic energy of electron = 1/2 m v²

= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²

= 40.95 x 10⁻¹⁹ J

Kinetic energy in electron comes from lose of electrical energy equal to

Ve where V is potential difference under which electron is accelerated and e is electronic charge .

V x e = kinetic energy of electron

V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹

V = 25.6 Volt .

Answer:

Explanation:

The point of destination is 141 north and 102 km east . Vectorially it is represented by unit vector as follows .

D = 102 i + 141 j

magnitude of D = √ ( 102² + 141² )

= √ ( 10404 + 19881)

= √ 30285

= 174 km

To go back to original position , ship should move on the following vector

D = -102 i - 141 j

Tan Ф = 141 / 102 = 1.38

Ф = 54⁰ , direction will be south of west . From north  east , angle will be

180 + 54 = 234⁰

Answer:

Force is too large for a person to exert.

Explanation:

m = Mass of child = 14 kg

u = Initial velocity of compartment = 64 mi/h = [tex]\dfrac{64\times 1609.34}{3600}=28.61\ \text{m/s}[/tex]

v = Final velocity of compartment = 0

t = Time taken by the compartment to stop = 0.05 s

Force is given by

[tex]F=ma\\\Rightarrow F=m\dfrac{v-u}{t}\\\Rightarrow F=14\times \dfrac{0-28.61}{0.05}\\\Rightarrow F=-8010.8\ \text{N}[/tex]

The person has to exert a force of [tex]-8010.8\ \text{N}[/tex] in the opposite direction to keep holding the child. This is a huge amount of force and cannot be done by a person's physical strength alone.

h max = 45.92 m

Given

1 kg ball

vo=initial velocity = 30 m/s

Required

Height

Solution

We can use the principle of energy conservation or parabolic motion

For parabolic motion

h max = vo²sin²θ/2g(θ= 90° for straight up)

For conservation energy :

KE₁=PE₂

1/2.m.vo² = m.g.h

We choose parabolic motion

h max = vo²sin²θ/2g

h max = 30²sin²90/2 x 9.8

h max = 45.92 m

Answer:

E. Some charges in the region are positive, and some are negative.

Explanation:

Electric potential is given as;

[tex]V = \frac{W}{Q}[/tex]

where;

W is the work done in moving a charge between two points which have a difference in potential

Q is quantity of charge in the given region

If the electric potential at a given point in the region is zero, then sum of the charges in the given region must be equal to zero. For the charges to sum to zero, some will be positive while some will be negative,.

Therefore, the correct statement in the given options is "E"

E. Some charges in the region are positive, and some are negative.

Answer:

Touch

Explanation:

Answer:

a) 13.7 m

b) 8.3 m

Explanation:

a)

       [tex]d = d_{1} + d_{2} = 11.0 m + 2.7 m = 13.7 m (1)[/tex]

b)

       [tex]x_{o} = 0 (2)[/tex]

       [tex]x_{f} = x_{1} - x_{2} = 11.0 m - 2.7 m = 8.3 m (3)[/tex]

      ⇒ Δx = xf - x₀ = 8.3 m  - 0 m =  8.3 m

Answer:

a)       v₂ = 13.20 m / s

Explanation:

To solve this exercise we will use the kinematic relations

Let's start with the Lioness.  Let's find the time to reach top speed

           v = v₀ + a t

as part of rest, its initial velocity is zero

           t = v / a

           t₁ = 21.0 / 2.57

           t₁ = 8.17 s

the total time is the acceleration time plus the time (t₂ = 25 s) that the maximum speed can withstand

          t = t₁ + t₂

          t = 8.17 +25.0 = 33.17 s

Now let's find out what distance the lioness travels in these times

during acceleration

         x = v₀ t + ½ a t²

         x = ½ a t²

         x₁ = ½ 2.57 8.17²

         x₁ = 85.77 m

during constant speed part

        x₂ = v t₂

        x₂ = 21.0 25.0

        x₂ = 525 m

therefore the total distance traveled is

        x = x₁ + x₂

        x = 85.77 + 525

        x = 610.77 m

a) the average speed of the gazelle

this must be the distance that the lioness travels minus the initial distance that separates the two animals (xo = 173 m) between the time taken

        v₂ = [tex]\frac{x -x_o}{t}[/tex]

        v₂ = [tex]\frac{610.77 - 173}{33.17}[/tex]

        v₂ = 13.20 m / s

b) in the attachment we can see a graph of the displacement of the two animals

Answer:

[tex]4.77\ \text{A}[/tex]

Explanation:

F = Magnetic force = 4.11 N

[tex]I_n[/tex] = Net current

[tex]I_2[/tex] = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

[tex]\theta[/tex] = Angle between current and magnetic field = [tex]65^{\circ}[/tex]

[tex]l[/tex] = Length of wires = 2.64 m

[tex]I[/tex] = Current in the other wire

Magnetic force is given by

[tex]F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}[/tex]

Net current is given by

[tex]I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}[/tex]

The current I is [tex]4.77\ \text{A}[/tex].

Answer:

I don't know lol good luck i guess

The average velocity of the car is 37.27 km/h.

The given parameters;

The average velocity of the car is calculated as follows;

[tex]average \ velocity = \frac{Total \ displacement}{Total \ time}[/tex]

The total displacement of the car is calculated as follows;

[tex]d = \sqrt{x^2 + y^2} \\\\d = \sqrt{100^2 \ + \ 50^2} \\\\d = 111.803 \ km[/tex]

The average velocity of the car is calculated as follows;

[tex]v = \frac{111.803}{3} \\\\v = 37.27 \ km/h[/tex]

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Answer:

take gravaiy and times it by 70m to find max v

Explanation:

ps i am olny 10 cool ya

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=[tex]9.81ms^{-2}[/tex]

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

[tex]v=u+at[/tex]

Using the formula and substitute the values

[tex]0=24.1-9.81t[/tex]

Where g is negative because motion of ball is against gravity

[tex]24.1=9.81t[/tex]

[tex]t=\frac{24.1}{9.81}=2.46s[/tex]

3.[tex]y=y_0+ut+\frac{1}{2}at^2[/tex]

Using the formula

[tex]y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2[/tex]

[tex]y=30.1 m[/tex]

4.Time of flight for red ball=3.77-2.9=0.87s

[tex]y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2[/tex]

[tex]y'=22.0m[/tex]

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

[tex]0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2[/tex]

[tex]0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2[/tex]

[tex]0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2[/tex]

[tex]0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t[/tex]

[tex]-2.86105=-2.851t[/tex]

[tex]t=\frac{2.86105}{2.851}=1.004 s[/tex]

Hence,  1.004 s  after the blue ball is thrown  are the two balls in the air at the same height.

Answer:

v = -19.6 m / s,    y =y₀ + v₀ t - ½ g t²

Explanation:

This is an exercise in kinetics in one dimension, let's take the upward direction as positive

           v = v₀ - g t

in this case as the body is released its initial velocity is zero and the acceleration is -g the sign indicates that it is directed downwards

           v = 0 -g t

           v = - 9.8 2

            v = -19.6 m / s

the sign indicates that the speed is down.

Galileo's equation is

            y =y₀ + v₀ t - ½ g t²

where i is the initial height, v₀ the initial velocity and -g the acceleration of the body

Explanation:

the state or quality of being efficient or able to accomplish something

Answer:

Explanation: relationship between the object and the observer's frame of reference.

Answer:

Illumination  = 200 fc

luminance =  2FL

Explanation:

given data

desired brightness = 100 fL

painted medium green reflectance = 50 percent

illumination = 4c

solution

we get here Illumination by using the formula that is

Illumination = luminance ÷ reflectance   ......................1

Illumination  = 100 fL ÷ 0.50

Illumination  = 200 fc

With light colored cream paint with a reflectance of 75%  the required illumination is 133 fc.

and

luminance will be

luminance = Illumination × reflectance = 4fc × 0.5 = 2FL

Cope up with changes is significant in everyone's life. One of the suitable way is to remind ourselves that we can withstand with the changes.

Reminding a thing that is important to anyone or ourselves until we do it a better way. Thinking past time and wasting time is just ridiculous. Because we must cope up with the current scenario and build up our mental strength.

Our lives are constantly impacted by changes. Change is a constant, uncontrolled aspect of life that has an impact on our work ethics and routines.

Anything that changes may have major or little repercussions. As a result, since the extent of change's first consequences is uncertain, we must refrain from discriminating against it.

Organizations can effectively use change as a tool. Without change, people would be hesitant to form fresh perspectives and ideas. They wouldn't be able to respond to daily obstacles and would become inert in their work and their roles.

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Answer:

distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

Explanation:

Given that;

mass of block m = 0.200 kg

distance travelled d = 0.796 m

time t = 2.00 s

m₂ = 0.400 kg

If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Now, using the second equation of motion;

d = ut + ([tex]\frac{1}{2}[/tex] × at²)

as the object started from rest, u=0

so, we substitute

0.796  = 0×2 + ([tex]\frac{1}{2}[/tex] × a(2)²)

0.796  = 0 + ([tex]\frac{1}{2}[/tex] × 4a)

0.796  = 2a

a = 0.796 / 2

a = 0.398 m/s²

using first equation of motion

[tex]V_{f}[/tex] = u + at

we substitute

[tex]V_{f}[/tex] = 0 + 0.398 × 2

[tex]V_{f}[/tex] = 0.796 m/s

now, average velocity is given as;

[tex]V_{avg}[/tex] = ( 0.796 m/s  + 0 ) / 2

[tex]V_{avg}[/tex] = ( 0.796 m/s  + 0 ) / 2

now, distance as the block moves in 2s will be;

D = [( 0.796 m/s  + 0 ) / 2 ] × 2

D = 0.796 m

Therefore, distance travelled by the block is 0.796 m

{ acceleration is independent of mass, so both the masses travel equal distance in 2 s }

The distance traveled by the object with the uniform motion can be given by the second equation of the motion.

The distance traveled by the big block with weight in 2 seconds is.

What is second equation of motion?

The distance traveled by the object with the uniform motion can be given by the second equation of the motion. It can be given as,

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

Given information-

The mass of the small block is 0.200 kg.

The total distance traveled by the block is 0.796.

Initial velocity of small block is zero.

Total time taken by the block to travel this distance is 2 seconds.

Put the values in the above equation as,

[tex]0.796=0\times 2+\dfrac{1}{2} a\times 2^2\\0.796=2a\\a=0.398[/tex]

Thus the acceleration of the small block is 0.398 meter per second.

Now the mass is doubled which is, 0.400 kg. As the acceleration does not depends on the mass, thus the acceleration for both cases is 0.398.

The velocity of the big block can be given as,

[tex]V=u+0.398\times2\\V=0+0.796\\V=0.796[/tex]

The velocity of the big block is 0.796.

The average velocity of the big block is given by,

[tex]V_{avg}=(\dfrac{0.796+0}{2} })\\V_{avg}=0.398[/tex]

The distance traveled by the object is the ratio of the velocity of the body to the time taken by it. Thus the distance traveled by the big block in 2 seconds is,

[tex]d={{0.398} \times2}\\d=0.796[/tex]

Thus the distance traveled by the big block with weight in 2 seconds is.

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A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.

I am unable to add the workings out, however please do message me and I will be able to provide you with them :)

Answer:

Conductivity, [tex]\sigma=500\ (\Omega-m)^{-1}[/tex]

Explanation:

Given that,

The resistivity of a material, [tex]\rho=2\times 10^{-3}\ \Omega-m[/tex]

We need to find the conductivity of the material.

We know that the reciprocal of resistivity is called conductivity.

[tex]\sigma=\dfrac{1}{\rho}\\\\\sigma=\dfrac{1}{2\times 10^{-3}}\\\\\sigma=500\ (\Omega-m)^{-1}[/tex]

So, the conductivity of the material is [tex]500\ (\Omega-m)^{-1}[/tex].

Answer:

Explanation:

Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.

[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]

Plug in all the values, and its algebra at this point

[tex]500=\frac{F*96}{12}[/tex]

6000 = 96F

F = 62.5 Newtons

The force which is being applied to the object with the power of 500 Watt is 62.5 N.

Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.

The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.

Therefore, the force of the object with power 500W can be calculated as:

P = f × d/ t

where, P = Power of the object,

f = Force applied,

d = distance travelled,

t = time taken to cover the distance

f = (P × t)/ d

f = (500 × 12) / 96

f = 6000/ 96

f = 62.5 N

Therefore, the force which is being applied to the object is 62.5N (Newton).

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Answer:

0.1835m/s

Explanation:

The formula for calculating the speed of wave is expressed as;

v = fλ

f is the frequency - The number of oscillations completed in one seconds

If 22 waves pass the boat every 60 seconds,

number of wave that passes in 1 seconds = 22/60 = 0.367 waves

Therefore the frequency f of the wave is 0.367Hertz

λ (wavelength) is the distance between successive crest and trough of a wave

λ = 0.5m

Substitute the given values into the formula

v = fλ

v = 0.367 * 0.5

v = 0.1835

Hence the speed of the waves is 0.1835m/s

Answer with Explanation:

We are given that

Length of pole(l0)=20 m

Length of garage(l)=10 m

a. We have to find the speed by which the Patrick is running.

We know that

[tex]l=l_0\sqrt{1-\frac{v^2}{c^2}}[/tex]

Substitute the values

[tex]10=20\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{10}{20}=\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{1}{2}=\sqrt{1-\frac{v^2}{c^2}}[/tex]

[tex]\frac{1}{4}=1-\frac{v^2}{c^2}[/tex]

[tex]\frac{v^2}{c^2}=1-\frac{1}{4}=\frac{3}{4}[/tex]

[tex]v^2=0.75c^2[/tex]

[tex]v=0.87c[/tex]

b.

l0=10 m

[tex]l=10\sqrt{1-\frac{3}{4}}[/tex]

[tex]l=5m[/tex]

Hence, the length of garage appear to him=5 m long.

Answer:

KE = ½mv²

KE = ½5kg×(20m/s)²

KE = ½5kg×400m²/s²

KE =5kg×200m²/s²

KE = 1000joules