A ball is dropped from a top of a 70 m tall building. How fast will the velocity be when it hits the
ground?

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:

The time that passes until the police catch the speeder is 82.6204 seconds.

Explanation:

A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.

The position is calculated by the expression:

x = x0 + v0*t + 1/2*a*t²

where:

First, let’s look at the police car’s equations of motion. In this case:

So: x = 50 m/s*t + 1/2*2 m/s²*t²

Now for the speeder’s car’s equations of motion you know:

So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

When the police catch the speeder they are both in the same position. So:

50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

Solving:

0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²

0= 3,000  + 55 *t + 1/2*t² - 50*t - 1*t²

0= 3,000  + 55 *t - 50*t - 1*t² + 1/2*t²

0= 3,000  + 5*t - 1/2*t²

Applying the quadratic formula:

[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]

x1= -72.6209

and x2= 82.6209

Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.

Time needed = t = 20.83 s

Given

car speed = 85 km/h

truck speed = 73 km/h

Required

the time it takes for the car to reach the truck

Solution

When the car reaches the truck, the distance between them will be the same

x car - 250 m = x truck

General formula for distance (d) :

d = v.t

So the equation becomes :

85t-250 = 73t

12t=250

t = 20.83 s

The maximum height : 81.63 m

Given

0.4 kg mass

vo = initial speed = 40 m/s

Required

the maximum height

Solution

We can use the law of conservation energy(ME=PE+KE) or use parabolic motion

For parabolic motion :

h max = (vo²sin²θ)/2g

θ = 90°(straight upward)

Input the value :

h max = (40²sin²90°)/2 x 9.8

h max = 81.63 m

Time needed = t = 20.83 s

Given

car speed = 85 km/h

truck speed = 73 km/h

Required

the time it takes for the car to reach the truck

Solution

When the car reaches the truck, the distance between them will be the same

x car - 250 m = x truck

General formula for distance (d) :

d = v.t

So the equation becomes :

85t-250 = 73t

12t=250

t = 20.83 s

Answer:

The balls velocity is 1 divided by 3

The velocity of the ball is 18.85 m/s.

The ball’s centripetal acceleration is 236.87 m/s².

The ball's centripetal force is  118.44 Newton.

Centripetal acceleration is a characteristic of an object's motion along a circular path. Centripetal acceleration applies to any item travelling in a circle with an acceleration vector pointing in the direction of the circle's center.

Given parameters:

length of the string: l = 1.5 meters.

Time interval = 60 seconds.

Total number of complete rotation = 120.

Hence, the velocity of the ball = 120×2π×1.5/60 m/s

= 18.85 m/s.

The ball’s centripetal acceleration = (velocity)²/ radius

= (18.85)²/1.5 m/s²

= 236.87 m/s²

The ball's centripetal force = mass × centripetal acceleration

= 0.5 × 236.87 Newton

= 118.44 Newton

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Answer:

Explanation:

Answer:

Distance, S = 440 meters.

Explanation:

Given the following data;

Initial velocity, u = 17m/s

Time, t = 20 seconds

Final velocity, v = 27m/s

To find the distance;

First of all, we would determine the acceleration of the truck.

Acceleration = (v-u)/t

Substituting the given values into the equation, we have;

Acceleration = (27 - 17)/20

Acceleration = 10/20

Acceleration = 0.5m/s²

Now, we would use the second equation of motion to find the distance traveled.

S = ut + ½at²

S = 17*20 + ½*0.5*20²

S = 340 + 0.25*400

S = 340 + 100

S = 440m

The equations of motion can be used to obtain the distance covered as 440 m.

We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;

v^2 = u^2 + 2as and v = u + at

v = final velocity

u = initial velocity

a = acceleration

s = distance

To obtain the acceleration;

27 = 17 + 20(a)

27 - 17 = 20a

a = 0.5 ms-2

Now, to obtain the distance;

v^2 = u^2 + 2as

v^2 - u^2/as = s

s = (27)^2 - (17)^2/2(0.5)

s = 440 m

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Explanation:

Moral duties are the duties performed by the people on the basis of humanity and moral values. The following are some of the moral duties :

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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]

Answer: 3. To give the object a negative charge we must give it more electrons, and this will increase its mass.

Explanation:

Suppose we have an object and we negatively charge it.

Then we are "adding" N electrons to the object.

Remember that the mass of an electron is:

m = 9.11*10^(-31) kg

Then if we add N electrons to an object of mass M, the new mass of the object will be:

Mass = M + N*9.11*10^(-31) kg

So we will have an (almost negligible) increase of the mass of the object.

(Something similar can happen if the object is positively charged, where we remove electrons, then the mass of the object decreases)

Then the correct option is:

3. To give the object a negative charge we must give it more electrons, and this will increase its mass.

The answer is "Infrared"

Hope this helps

Answer:

x-rays

Explanation:

Explanation:

1.

We use the equation

h = [tex]\frac{gt^2}{2}[/tex], where

h is the height traveled,

g is the acceleration due to gravity and

t is the time taken to reach height h.

We can now calculate t to be

[tex]\sqrt{\frac{2*1.2 m}{9.81 m/s^2} }[/tex]

= 0.495 s

Let v be the initial velocity of the player.

The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.

v = 9.81 m/s^2 x 0.495 s = 4.85 m/s

2.

The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is

4.85 m/s / 0.3 s = 16.2 m/s^2

Frequency = (speed) / (wavelength)

Frequency = (3 x 10⁸ m/s) / (124 x 10⁻⁹ m)

Frequency = 2.42 x 10¹⁵ Hz

Answer:

Explanation:

Kinetic energy ( KE ) = 1/2 m v²

= m²v² / 2 m = p² / 2m where p is momentum

KE = p² / 2m

p² = 2m KE

KE is constant

p is proportional to mass

So car having higher mass will have higher momentum .

p₁ = √ ( 2 m x KE )

p₂ = √ ( 6 m x KE )

p₂ - p₁ =  √ ( 6 m x KE ) - √ ( 2 m x KE )

= √KE m ( √6 - √2 )

Kinetic energy ( K.E )

[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]

where p is momentum

[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]

KE is constant

p is proportional to mass

So car having higher mass will have higher momentum .  

[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]

The difference is shown above.

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Answer:

a) 4.0816s

b) -9.8 ms^-1

c) 81.63265m

Answer:

[tex]a=20.14\ m/s^2[/tex]

Explanation:

The time period of the simple pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is the length of the pendulum

g is the acceleration due to gravity

We have,

T = 1.4 s, l = 1 m

So,

[tex]T^2=\dfrac{4\pi^2 l}{g}\\\\g= \dfrac{4\pi^2 l}{T^2}\\\\g= \dfrac{4\pi^2 \times 1}{(1.4)^2}\\\\g=20.14\ m/s^2[/tex]

So, the acceleration due to gravity of that planet is [tex]20.14\ m/s^2[/tex].

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

Answer:

the number of carousel conveyors that an operator can operate without any idle time is 5

Explanation:

Given the data in the question;

first we express the equation for number of carousel conveyors that can be operated by an operator;

n' = [tex]\frac{(a + t)}{( a + b)}[/tex]

where a is the concurrent activity time ( 0.25 minute )

b is the independent operator activity time

t is the independent machine activity time( 1 )

Now independent activity time is zero as the operator is not performing any inspection or packaging tasks.

So time taken for the operator to retrieve the finished item at the end of the process is the concurrent activity and independent machine activity time, the conveyor rotation time of each item

so

we substitute

0.25min for a, 1 for t and 0min for b

n' = [tex]\frac{(0.25min + 1min)}{( 0.25min+ 0 min)}[/tex]

n' = 1.25 min / 0.25

n' - 5

Therefore, the number of carousel conveyors that an operator can operate without any idle time is 5

Answer:

blue

Explanation:

nswer the following about two objects, A and B, whose motion produced ihe following ... "A" starts with a greater (t) position. Since they ... since you are moving away from the origin

Answer:

73.3m/s

Explanation:

We can find the velocity of the player.

Momentum = mass * velocity

Given

Mass = 0.15kg

Momentum = 11kgm/s

Get the velocity

Velocity = Momentum/Mass

Velocity = 11/0.15

Velocity = 73.3m/s

Hence the velocity of the player is 73.3m/s

Answer:

Explanation:

Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .

a )

Given,

Normal force N = 400 N

Force of friction = 300 N

μ = coefficient of static friction = ?

Putting the values ,

300 = 400 μ

μ = .75

b )

Normal force N = 400 N

Force of friction = 200 N

μ = coefficient of kinetic  friction = ?

Putting the values ,

200 = 400 μ

μ = .50

c ) see attached file .

Answer:

v=0.14c

Explanation:

Answer:

making sense of a 3-d world from 2-d data

Explanation:

Answer:

Explanation:

individual and then net

hope that helps I could be wrong about this one though

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). (a) in the same direction as , (d) lower

ii). (f) opposite the direction of, (g) higher

Explanation:

An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.

An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.

Answer: Family

Explanation:

Solution :

Length of the plastic rod , L = 1.6 m

Total charge on the plastic rod , Q = [tex]$-9 \times 10^{-8}$[/tex] C

The rod is divided into 8 pieces.

a). The length of the 8 pieces is , [tex]$l=\frac{L}{8}$[/tex]

                                                         [tex]$=\frac{1.6}{8}$[/tex]

                                                         = 0.2 m

b). Location of the center of the piece number 5 is given as : 0 m, -0.09375 m, 0 m.

c). The charge q on the piece number 5 is given as

[tex]$q=\frac{Q}{L}\times l$[/tex]

[tex]$q=\frac{-9 \times 10^{-8}}{1.6}\times0.2$[/tex]

 = [tex]$-1.125 \times 10^{-8}$[/tex] C

d). WE approximate that piece 5 as a point charge and we need to find out the field at point A(0.7 m, 0, 0) only due to the charge.

We know,  the Coulombs force constant, k = [tex]$8.99 \times 10^9 \ N.m^2/C^2$[/tex]

So the X component of the electric field at the point A  is given as

[tex]$E_x = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \cos \frac{187.628}{0.70625}$[/tex]

     = -126.15 N/C

The Y component of the electric field at the point A is

[tex]$E_y = 8.99 \times 10^9 \times 1 \times 10^{-8} \ \sin \frac{187.628}{0.70625}$[/tex]

    = -16.93 N/C

Now since the rod and the point A is in the x - y plane, the z component of the field at point A due to the piece 5 will be zero.

∴ [tex]$E_z=0$[/tex]

Thus, [tex]$E= <-126.15,-16.93,0>$[/tex]