What is the car's average velocity (in m/s) in the interval between t = 0.5 s
to t = 2 s?

Answer:

a full length of image of a distance tall building can definitely be seen by using telescope

Answer:

Explanation:

We shall apply newton's laws formula

a )

initial velocity in upward direction u = 10 m/s

acceleration due to gravity g = 9.8/ m .s²

Let h be the maximum height where v = o

v² = u² - 2gh

0 = 10² - 2 gh

h = 10² / 2g

= 10² /  2  x 9.8

= 5.10 m

Since the ball was thrown from height of 100 m , total maximum height of ball

= 100 + 5.10

= 105.10 m

Let t be the time taken

v = u - gt

0 = 10 - gt

t = 10 / 9.8

= 1.02 s

b )

when h = 50 on its way downwards , velocity

v² = u² + 2 g s  

v² = 0 + 2 x 9.8 x ( 105.10 - 50 )

[  distance travelled by ball at this point from top = 105.1 - 50 = 55.10 ]

v = 32.86 m / s

Let us find out final velocity of touching the ground . For it distance travelled = 105.10

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 105.1

v = 45.39 m /s

Now velocity at h = 50 is 32.86

velocity at h = 0 is 45.39

time taken to travel fro h = 50 to h = 0

v = u + gt

45 .39 = 32.86 + 9.8 x t

t =  1.28 s .

Answer:

The torque of the wheel is 0.92 N-m.

Explanation:

Given that,

Radius of wire = 3 cm

Angular acceleration = 92.00 rad/s²

Suppose, Mass of wheel is 4 kg and radius is 5 cm. find the torque of the wheel.

We need to calculate the torque of the wheel

Using formula of torque

[tex]\tau=I\alpha[/tex]

[tex]\tau=MR^2\times\alpha[/tex]

Where, M = mass of wheel

R = radius of wheel

[tex]\alpha[/tex] = angular acceleration

Put the value into the formula

[tex]\tau=4\times(5\times10^{-2})^2\times92[/tex]

[tex]\tau=0.92\ N-m[/tex]

Hence, The torque of the wheel is 0.92 N-m.

Answer:

57,9 m/s

Explanation:

A train is moving at a constant speed on a surface inclined upward at 15.0° with the horizontal and travels 300 meters in 5 seconds. The horizontal velocity of the train at the end of 3 seconds is 100m/s.

The first formula we will use is:

v = d/t

300m ÷ 5s  = 60m/s

The last formula we will use is:

vh = v·cosx

60m/s·cos15.0°

60m/s · 0,965  = 57,9 m/s

Therefore, the answer is 57,9 m/s.

Hope this helped!

57,9 m/s

velocity formula

v = d/t

= 300m/5s

= 60m/s

horizontal velocity formula

vh = v·cosx

= 60m/s×cos15°

= 60m/s×0,965

= 57,9 m/s

Answer:

32

Explanation:

The average speed of the car for the remaining is  32 miles/h.

The average speed of any moving object is the ratio of the total distance covered and the total time taken to cover that distance.

Average Speed S= distance /time

Given is a sailboat took 2.5 hours to cover 1/4 of a journey. Then, it covered the remaining 144 miles in 3.5 hours.

If x be the total distance covered during journey, then

3/4 x = 144

x = 192 miles

The total time taken for journey is 2.5 +3.5 = 6 h

The average speed for the whole journey is

Avg speed S= 192 / 6

S = 32 miles/h

Thus, the average speed of the car for the remaining is  32 miles/h.

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Answer:

412.5

Explanation:

all you have to do is multiply .15

Answer:

Explanation:

s(t) = -16t2 + 8√t

A )

Average velocity

s(t) / t = (-16t2 + 8√t)/t

A(t)= -16t +  8 / √t

average velocity of the ball for t near 4.

A(t) =  -16t +  8 / √t

          Lt t⇒4

B )

Distance covered in 4 s

-16t2 + 8√t

= - 16 x 16 + 8 x 2

= - 240

Distance covered in 5 s

= - 16 x 25 + 8 √5

= -400 + 17.88

= -382.12

distance covered in duration from 4 to 5 sec

= -142.12

velocity = - 142.12 / 1 = - 142.12 m /s

Distance covered in 4.5 s

= -16 x 4.5² + 8√4.5

= -324 + 16.97

= -307

distance covered during 4 to 4.5

= 67

velocity during 4 to 4.5

= -67 / .5

- 134 m /s

distance covered in 4.05 s

-16 x 4.05² + 8√4.05

-262.44 + 16.1

-246.34

distance covered during 4 to 4.05

= -6.34

velocity during 4 to 4.05

= -6.34 / .05

- 126.8 m /s

C )

Instantaneous velocity at t = 4

= - 120 m /s

(A) As 't' approaches to 4s, the formula of average velocity is

[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]

(B) The average velocity for the time interval starting at t = 4 with a duration of 1 second is [tex]142.11\,m/s[/tex].

The average velocity for the time interval starting at t = 4 with a duration of 0.5 seconds is [tex]-134.058\,m/s[/tex]

The average velocity for the time interval starting at t = 4 with a duration of 0.05 seconds is [tex]-126.8\,m/s[/tex].

(C) The instantaneous velocity of the ball at 4s is [tex]-126\,m/s[/tex].

The answers are explained as follows.

Given that the distance is a function of time.

[tex]s(t)=-16\,t^2\,+\,8\sqrt{t}[/tex]

(A) The average velocity can be given by,

[tex]v_{avg}=\frac{-16\,t^2\,+\,8\sqrt{t} }{t} =-16t+ \frac{8}{\sqrt{t}}[/tex]

As 't' approaches to 4s, the formula becomes;

[tex]v_{avg} _{|(t\rightarrow4)}=Lim_{|(t\rightarrow4)}[-16t+ \frac{8}{\sqrt{t}} ][/tex]

(B) We know that the average velocity or in this case speed is the total distance by the total time taken.

Distance covered in 4 s can be found by putting [tex]t=4s[/tex] in the distance formula.

Distance covered in 5 s can also be found by the same method

Therefore, the distance covered in from 4 to 5 seconds is;

Distance covered in 4.5 s is given by,

Therefore, the distance covered in 4 to 4.5 seconds is;

Distance covered in 4.05 s is given by,

Therefore, the distance covered in 4 to 4.05 seconds is;

(C) The instantaneous velocity of the ball can be found by differentiating the function [tex]s(t)[/tex].

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Answer: Procrastination, Defiance, Laziness

Explanation:

Answer: impulsiveness

Explanation:

Answer:

erm they all should have same numbers of protons

Answer:

They each have the same number of nuclear particles.

Explanation:

What is true about all uranium atoms? They each have the same number of nuclear particles. They each have the same number of neutral particles. They each have the same number of neutrons . Uranium is an element that is often used in nuclear power plants. I hope this helped :)

Answer:

A. 0.025inches= 1m

So 75in= 0.025*75= 1.88m

B 3.45*10^6yrs= 3.45E6* 365x 86900s

= 1.09*10^8s

C. 62ft/day= 62* 0.3048/86900= 2.1*10^-4

D. 2.2*10^4mi² x ( 1609.3)²

= 6.1*10^7m²

Answer:

Its B

Explanation:

the force of the ball is being kicked from the back side of the ball will causes a pushing motion and the ball will move forward

Answer:

Explanation:

Using the equation of motion to get the initial velocity as it leaves the ground. According to the equation of motion;

v² = v0²+2as where;

v is the final velocity

v0 is the initial velocity

a is the acceleration due to gravity

s is the height.

At the maximum height, the final velocity v = 0m/s

Given s = 0.500m and a = g = -9.81m/s² (acceleration due to gravity is negative due t the upward movement of the flea)

0² = v0² - 2(9.81)(0.500)

0 =  v0² - 9.81

- v0² = -9.81

v0 = √9.81

v0 = 3.13m/s

Hence the initial velocity v0 of the flea as it leaves the ground is 3.13m/s (to three significant figures)

Explanation:

1) F₁ lies in a plane perpendicular to the xy plane, 60° from the x axis.  The angle between F₁ and the +z axis is 30°.  Therefore, the vector is:

<F₁> = 20 (sin 30° cos 60° i + sin 30° sin 60° j + cos 30° k)

<F₁> = 20 (¼ i + ¼√3 j + ½√3 k)

<F₁> = 5 i + 5√3 j + 10√3 k

F₂ is in the xy plane.  Its slope is -24/7.  The vector is:

<F₂> = 12 (-⁷/₂₅ i + ²⁴/₂₅ j + 0 k)

<F₂> = -3.36 i + 11.52 j

F₃ is parallel to the +x axis.  The vector is:

<F₃> = 17 (i + 0 j + 0 k)

<F₃> = 17 i

F₄ is parallel to the -z axis.  The vector is:

<F₄> = 15 (0 i + 0 j − k)

<F₄> = -15 k

F₅ is in the xy plane.  It forms a 15° angle with the -y axis.  The vector is:

<F₅> = 9 (-sin 15° i − cos 15° j + 0 k)

<F₅> = -9 sin 15° i − 9 cos 15° j

The resultant vector is therefore:

<F> = (5 − 3.36 + 17 − 9 sin 15°) i + (5√3 + 11.52 − 9 cos 15°) j + (10√3 − 15) k

<F> = 16.31 i + 11.49 j + 2.32 k

2) Sum of forces at point B in the x direction:

∑F = ma

Tbc cos 40° − ¹⁵/₁₇ Tab = 0

Tbc cos 40° = ¹⁵/₁₇ Tab

Tbc = 1.15 Tab

Sum of forces at point B in the y direction:

∑F = ma

Tbc sin 40° + ⁸/₁₇ Tab − mAg = 0

Tbc sin 40° + ⁸/₁₇ Tab = (2 kg) (10 m/s²)

(1.15 Tab) sin 40° + ⁸/₁₇ Tab = 20 N

1.21 Tab = 20 N

Tab = 16.52 N

Tbc = 19.02 N

Sum of forces at point C in the x direction:

∑F = ma

Tcd sin 25° − Tbc cos 40° = 0

Tcd sin 25° = Tbc cos 40°

Tcd = 1.81 Tbc

Tcd = 34.48 N

3(a) When the crane is on the verge of tipping, the center of gravity is directly over point F.  Relative to point A:

3.7 m = [ (390 kg) (0.9 m) + (90 kg) (9 m cos θ + 1.7 m) + (80 kg) (9 m cos θ + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + 810 kgm cos θ + 153 kgm + 720 kgm cos θ + 232 kgm

1336 kgm = 1530 kgm cos θ

θ = 29.17°

3(b) 3.7 m = [ (390 kg) (0.9 m) + (90 kg) (x + 1.7 m) + (80 kg) (x + 2.9 m) ] / (390 kg + 90 kg + 80 kg)

2072 kgm = 351 kgm + (90 kg) x + 153 kgm + (80 kg) x + 232 kgm

1336 kgm = (170 kg) x

x = 7.86 m

4) Find the lengths of the cables.

Lab = √((2 m)² + (3 m)² + (5 m)²)

Lab = √38 m

Lac = √((2 m)² + (3 m)² + (5 m)²)

Lac = √38 m

Lde = √((2 m)² + (3 m)²)

Lde = √13 m

Sum of forces in the x direction:

∑F = ma

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Sum of forces in the y direction:

∑F = ma

2/√38 Fab − 2/√38 Fac = 0

Fab = Fac

Sum of forces in the z direction:

∑F = ma

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

Sum of moments about the y-axis:

∑τ = Iα

(3/√38 Fab) (5 m) + (3/√38 Fac) (5 m) + (3/√13 Fde) (2 m) − (mg) (2 m) = 0

Substitute Fab = Fac and simplify:

6/√38 Fab + 3/√13 Fde − mg = 0

30/√38 Fab + 6/√13 Fde − 2mg = 0

Double first equation:

12/√38 Fab + 6/√13 Fde − 2mg = 0

Subtract from the second equation:

28/√38 Fab = 0

Fab = 0

Fac = 0

Solve for Fde:

3/√38 Fab + 3/√38 Fac + 3/√13 Fde − mg = 0

3/√13 Fde = mg

3/√13 Fde = (1.7 kg) (10 m/s²)

Fde = 20.43 N

Solve for Rx:

-5/√38 Fab − 5/√38 Fac − 2/√13 Fde + Rx = 0

Rx = 2/√13 Fde

Rx = 11.33 N

◈ A particle is moving along the x-axis according to the equation

____________________________

✪ Position at t = 3s :

➝ S = 4 + 6t - 2t²

➝ S = 4 + 6(3) - 2(3)²

➝ S = 4 + 18 - 18

➝ S = 4m

✪ Instantaneous velocity at t = 3s :

➝ v = dx/dt = d(4 + 6t - 2t²)/dt

➝ v = 6 - 4t

➝ v = 6 - 4(3)

➝ v = 6 - [tex]\sf{12}[/tex]

➝ v = -12m/s

✪ Instantaneous acc. at t = 3s :

➝ a = dv/dt = d(6 - 4t)/dt

➝ a = -4m/s²

[Acceleration does not depend on time]

Answer:

Empathetic people try to look at things from different  perspectives

People with low empathy often are unable to understand others perspective and involve in unproductive arguments.

sympathy is when you feel sorry for the other person and there is  no urge, no impelling reason to be  feel others feelings.

​

Explanation:

Answer:

Explanation:

1 psi = 6894.76 Pa

P = 184 psi = 12.686 x 10⁵ Pa .

Temperature T = 95⁰F = 35⁰C= 308 K

volume V = 18 ft³ = 18 x 0.0283168 m³

= .51 m³

From the gas law

PV/RT = n where n is mole of gas

= 12.686 x 10⁵ x  .51  / 8.31 x 308

= 252.78 gm mole

= 252.78 x 32 gm

= 8.08896 kg

= 2.20462 x 8.08896 lb

= 17.833 lb

= 17.833 / 32 lbf

= .5573 lbf

weight of tank = 150 lbf

Total weight = 150 + .5573

= 150.5573 lbf .

Answer:

3.96*10^-3 m

Explanation:

Given that

Amplitude, A = 10000 mm/s² = 10 m/s²

Frequency, f = 8 hz

the maximum acceleration of a simple harmonic motion is given as

|a|(max) = Aw², where

A = amplitude of the simple harmonic motion

w = angular frequency of the simple harmonic motion

Remember, w = 2πf, so

w = 2 * 3.142 * 8

w = 50.272 rad/s

Rearranging the formula, and making a the subject, we have

A = |a|(max) / w², so that

A = 10 / 50.272²

A = 10 / 2527.27

A = 0.00396 m

A = 3.96*10^-3 m

Answer:

Explanation:

Velocity is the change in displacement of a body with respect to time.

Velocity = Displacement/Time

Given velocity = 5m/s

Displacement = 1.5 km

Displacement = 1.5km * 1000 = 1500m

From the formula, Time = Displacement/Velocity

Time = 1500/5

Time = 300 secs

Hence the time it will take to travel 1.5km is 300secs

Explanation:

Hey there!!

Here,

Initial velocity (u) = 6m/s.

Acceleration (a) = 1.7m/s^2.

Time (t) = 4.2s.

final velocity (v) = ?

We have,

[tex]a = \frac{v - u}{t} [/tex]

Putting their values,

[tex]1.7 = \frac{v - 6}{4.2} [/tex]

7.14 = v - 6

v = 7.14 + 6

Therefore, the final velocity is 13.14 m/s.

Hope it helps....

Answer: Mathematical Model

Explanation:

Took the test

Answer:

a precise, testable statement of what the reseatcher predict will be the outcome of the study

Explanation:

Answer:

supposition or proposed explanation

Explanation:

its made on the basis of limited evidence as a starting point for further investigation.

Answer:

HEY THERE! HERE'S YOUR ANSWER!

Explanation:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

HOPE THIS HELPS YOU BUDDY!

Answer:

n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

Explanation:

Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.

In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.

              [tex]a_{t}[/tex] = (v₂-v₁)/Δt

In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.

                    [tex]a_{c}[/tex] = v2/R

In the general case, both the module and the address change

             a = Ra (  a_{t}^2 +   a_{c}^2)  

The velocity of an object changes if it undergoes uniformly acceleration

motion by considering if it is:

This is the rate of change of an object's position with time. If the object has

positive constant acceleration, the slope goes upward while if it is a

negative constant acceleration, the slope goes downward.

The direction doesn't change as a result of the uniform speed and

direction that is involved.

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Answer:

Travelling fastest: when the football leaves the hands of the quarterback and when the receiver gets it

Travelling slowest: at the very top of then arched trajectory

Explanation:

Notice that this example can be analyzed with a two dimensional pattern as in the launching of a cannonball, where we study separately the velocities and acceleration acting vertically and horizontally.

The football is thrown at an agle with the horizontal with an initial velocity that is decomposed in the vertical and in the horizontal axes.

The Horizontal movement of the football is that of an object with constant velocity (that of the horizontal component of the initial velocity imparted by the quarterback) .

The vertical movement of the football is that of an object moving in an accelerated fashion, with constant acceleration due to the gravitational field, and with an initial velocity opposite to that constant acceleration. The initial velocity is that of the vertical component of the initial velocity imparted by the quarterback.

AT every point on the path of the ball, its velocity can be calculated by the vector addition of the horizontal component (with constant velocity as we discussed above) and the vertical component (this velocity is changing since its is under accelerated motion) of the object's velocity at any given time through the path.

The maximum velocity of the football will be at the point where the two components are at their maximum (that is when the football leaves the hand of the thrower, and when it gets to the hands of the receiver.

While the minimum is going to be when the vertical component of the velocity is at is minimum (zero) at the top of the arched trajectory when the football changes vertical direction and starts heading downwards.

Answer:

Data

Explanation:

Answer:

Determining how data will be gathered

Explanation:

As a scientist considers the use of a particular set of experimental procedures to carry out his research, an important consideration is the method of data collection.

The experimental procedures chosen must be those that make the process of data collection much easier, efficient and lead to collection of reliable data.

Hence determining the method of data collection is very important when designing a set of experimental procedures.