The driver of a car traveling at 21.6 m/s applies the brakes and undergoes a constant deceleration of 1.86 m/s 2 . How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of 0.35 m

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:

No

Explanation:

For infinite speed to be achevied, one must have no sink of energy to spend. The source of entropy in this example, is the tires hitting the surface, producing heat and friction. Not to mention that you'd still need fuel to start the car, and an infinite tunnel or track, which would be impossible and speed up to process of energy loss through entropy quicker.

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:

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:

Stress, in physical sciences and engineering, force per unit area within materials that arises from externally applied forces, uneven heating, or permanent deformation and that permits an accurate description and prediction of elastic, plastic, and fluid behaviour.

I hope it's helpful!

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.

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:

Touch

Explanation:

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

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:

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

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.

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 :)

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:

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

The choices that represent the step in the oil formation process are:

Layers of sediment cover dead remains.

Organisms die and sink to the ocean floor.

Heat and pressure change dead remains.

Oil and gas should be created from organic material that is deposited with respect to the sediments on the seabed and after that, it brokes down and then transformed over millions of years. The sediment layers should be covered and dead. The organisms die and then they sink to the ocean. The change to the heat and pressure should remain dead.

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Answer: 2, 3, and 4.

Organisms die and sink to the ocean floor.

Heat and pressure changes dead remains.

Layers of sediment cover dead remains.

Explanation:

I just took the test

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:

A) v(t) = (32 - 1.6t) m/s

a(t) = -1.6 m/s²

B) t = 20 seconds

C) 320 m

D) t = 5.85 seconds going up and 34.14 seconds going down

E) 40 seconds

Explanation:

Height is given by the equation;

S(t) = 32t - 0.8t²

A) Velocity after time t is gotten from first derivative of the distance.

Thus;

v(t) = dS/dt = 32 - 1.6t

Acceleration at time t is gotten from derivative of the velocity.

Thus;

a(t) = d²S/dt² = -1.6 m/s²

B) At highest point, velocity is zero.

Thus;

32 - 1.6t = 0

1.6t = 32

t = 32/1.6

t = 20 seconds

C) To find how high the rock goes, it means we are looking for maximum height.

This will be at t = 20 seconds.

Thus;

S(20) = 32(20) - 0.8(20)²

S(20) = 640 - 320

S(20) = 320 m

D) we want to find the time it will take to reach half its maximum height.

since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160

Thus;

160 = 32t - 0.8t²

0.8t² - 32t + 160 = 0

Using quadratic formula, we will get;

t = 5.85 seconds going up and 34.14 seconds going down

E) time the rock is aloft = twice the time it took to reach maximum height.

Thus; t_aloft = 2 × 20 = 40 seconds

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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Answer:

t = 25.0 s

Explanation:

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

       [tex]v_{o} = 84.1 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 23.4 m/s (2)[/tex]

      [tex]v_{f} = 15.8 km/h*\frac{1h}{3600s} *\frac{1000m}{1km} = 4.4 m/s (3)[/tex]

       [tex]a = \frac{(4.4m/s)^{2} - (23.4m/s)^{2}}{2*350m} = -0.76 m/s2 (4)[/tex]

       [tex]t =\frac{v_{f}-v_{o}}{a} = \frac{(4.4m/s)-(23.4m/s)}{-0.76m/s2} = 25.0 s (5)[/tex]

The velocity of the freely falling object reached after 4.0 s is equal to 39.2 m/s.

A relation can be established between the velocity of the body and acceleration of the body when the body is moving along a straight line with uniform acceleration. The distance traveled by the body in a specific time has expressed a set of equations called equations of motion.

The first equation of motion can be expressed as follows:

v = u + at

Where u is the initial velocity, v is the final velocity and a is the acceleration and t is the time of the object.

Given, the acceleration of the falling object, g = 9.8 m/s²

The initial velocity of the object, u = 0, and the time = 4 sec

The velocity of the object after time, t = 4 sec.

v = (0) + (9.80)× (4)

v = 39.2 m/s

Therefore, the velocity of the falling object will be 39.2 m/s after 4 seconds.

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Answer:

KE = ½mv²

KE = ½5kg×(20m/s)²

KE = ½5kg×400m²/s²

KE =5kg×200m²/s²

KE = 1000joules

Answer:

   V_{2G} = - 10 m / s

Explanation:

This is an exercise on relative velocities in one dimension, let's use the subscript 1 for the girl, the 2 for the but, the index 3 for the conveyor belt and the index G for the Earth

we will consider the velocity is positive to the right.

They indicate the speed of the belt with respect to the ground = 2 m/s

   the dog speed with respect to the belt = - 8 m / s

Dog speed is requested with respect to the Earth

            [tex]V_{2G} = V_{23} - V_{3G}[/tex]

notice that the inner subscripts cancel out.

We calculate

            = -8 - 2

            V_{2G} = - 10 m / s

the negative sign indicates that the dog moves to the left

Answer:

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

Explanation:

ps i am olny 10 cool ya

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].