the equation of motion is given for a particle when s is in meters and t is in seconds. Find the acceleration after 4.5 seconds s

We are given:

Mass of object (m) = 12 kg

acceleration (a) = 6 m/s²

Solving for the Force:

From newton's second law of motion:

F = ma

replacing the variables

F = 12*6

F = 72N

The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.

It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.

As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.

The momentum of the cyclist = 88 × 2.4

                                                 = 211.0 kgm/s

Thus, the momentum of the cyclist would be  211.0 kgm/s.

To learn more about momentum from here, refer to the link given below;

brainly.com/question/17662202

#SPJ2

Answer:

they don't strech so they tear a muscle when they perform

Explanation:

Answer: KE= 324,000

Explanation: I hope that this helps! -_-

Answer:

324,000

Explanation:

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Explanation:

From the question we are told that

  The electric field is  [tex]E(r)_1 =  [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]

    Now when the outer diameter is doubled, the radius(b) is also doubled

So

    [tex]E(r)_2 =  [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]

Now

      [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]

=>   [tex]\frac{E(r)_2}{E(r)_1}  =  \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]

=>  [tex]\frac{E(r)_2}{E(r)_1}  =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]

Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]

Answer:

Vbx = 25 * cos 340 = 23.5 knot        x-component of boats speed

Vrx = 20 cos 175 = -19.9 knot        x-component of rivers speed

Vx = 3.58 knot      x-component of boat and river speed

Vby = 25 sin 340 = -8.55 knot    y-component of boat speed

Vry = 20 sin 175 = 1.74 knot    y-component of river speed

Vy = -6.81 knot     y-component of boat and river speed

V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots

Answer:

I think it's = 12 seconds

Explanation:

the formula for speed is:

speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:

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

(sub the numbers in the formula)

distance=1200m, speed=100m/s

time=[tex]\frac{1200}{100}[/tex]

=12 seconds

Answer:

Mg (atomic number 12)

K (atomic number 19)

Explanation:

The size of an atom is estimated in terms of its atomic radius.

The atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance (d) between two nuclei in the solid - state of metals.

Cl is further right of Mg in the third period

K is below Na in the first group

Answer:

Explanation:

the distance have the following relation:

d = (1/2)gt2

D=32.0 m

t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s

it take 2.56s from the glasses to hit the ground

when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s

x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m

the pen only travel 1.54m

so the pen is above the ground 32.0m - 1.54m = 30.46m

The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.

Height is a numerical representation of the distance between two objects or locations on the vertical axis.

The height can refer to a physical length or an estimate based on other factors in physics or common use. |

The given data in the problem is;

h is the height from the top of a stadium = 32.0 m

t is the time period when the pen is dropped later =  2.00 s

x is the height above the ground

a is the air resistance. a = -g = -9.81 m/s²

From the second equation of motion;

[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]

When the glasses fall to the ground, the pen only travels a short distance;

[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]

So the pen travel the distance;

[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]

The pen above the ground is found as;

[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]

Hence the pen is 30.46m above the ground. when the glasses hit the ground.

To learn more about the height refer to the link;

https://brainly.com/question/10726356

Answer:

Victor will always be able to select 4 of those cards with the following property

Explanation:

Number of trading cards = 100

victor selects 21 cards

let the 4 cards be labelled : A,B,C and D

The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P

let the two pairs be : ( A + B ) and ( C + D )

note: average power of each pair = P  and this shows that

( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property

we have to check out the possible choices of two cards out of 21 cards yield distinct sums.

= C(21,2)=(21x20)/2 = 210.

from the question the number of distinct sums that can be created using 101  through 200 is < 210 .

hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

Answer:

6.7 seconds

Explanation:

d=(1/2)at^2

equation

1000=(1/2)45t^2.

substitute

2000=45t^2.

multiply by 2 for both sides

44.44=t^2.

divide both sides by 45

6.7=t

take the square root of both sides

Answer:

The one with the faster velocity is the one with a velocity of -10m/s

Answer:

Explanation:

To find the force given the mass , velocity and time can be found by using the formula

[tex]f = \frac{m \times v}{t} \\ [/tex]

where

m is the mass

v is the velocity

t is the time

From the question

m = 1500 kg

v = 30 m/s

t = 3 s

We have

[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

We are given:

initial velocity (u) = 12 m/s

final velocity (v) = 0 m/s

time taken (t) = 15 seconds

acceleration (a) = a m/s²

Solving for acceleration:

from the first equation of motion

v = u + at

replacing the variables

0 = 12 + (a)(15)

0 = 15a + 12

a = -12 / 15

a = -4 / 5 m/s²

Answer:

20000

Explanation:

Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000

Answer:

milliliters (ml)

Explanation:

millileters is the correct measurement for liquids

Answer:

Explanation:

mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water

A = 3.14 x .005²

= 785 x 10⁻⁷ m²

mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg

momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .

Rate of change of momentum  = 706.5 x 10⁻⁴

Let force be F

F - mg = 706.5 x 10⁻⁴

F = mg + 706.5 x 10⁻⁴

F =  23.55 x 10⁻⁴  x 9.8 + 706.5 x 10⁻⁴

= 937.3 x 10⁻⁴ N .

Answer:

h = 2 m

Explanation:

Mechanical advantage of a ramp is given by :

MA = length of incline/height of incline

Length of the ramp, l = 12 m

MA = 6

We need to find the height of the ramp.

So,

height of ramp = length of incline/MA

h = 12/6

h = 2 m

So, the height of the ramp is 2 m.

Answer:

108 metres

Explanation:

Given

[tex]Force = 900N[/tex]

[tex]Mass = 600kg[/tex]

[tex]Time = 12s[/tex]

Required

Determine the distance moved

First, we need to determine the acceleration.

[tex]Force = Mass * Acceleration[/tex]

[tex]900N = 600kg * a[/tex]

Solve for a

[tex]a = 900/600[/tex]

[tex]a = 1.5m/s^2[/tex]

Next, we determine the distance using:

[tex]S = ut + \frac{1}{2}at^2[/tex]

Since it starts from rest,

[tex]u = 0[/tex]

[tex]t = 12[/tex]

[tex]a = 1.5[/tex]

So:

[tex]S = 0 * 12 + \frac{1}{2} * 1.5 * 12^2[/tex]

[tex]S = \frac{1}{2} * 1.5 * 144[/tex]

[tex]S = \frac{1}{2} * 216[/tex]

[tex]S = 108m[/tex]

Answer:

His weight would be zero on the scale i.e he is weightless at that instance.

Explanation:

weight = mg

where m is the mass of the object, and g is the acceleration of gravity.

⇒ 810 = mg

During free fall, the weight of an object can be determined by:

W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)

where a is the acceleration of the object.

But since James fall at the acceleration of gravity, then:

g = a

mg = ma = 810 N

So that;

W = 810 - 810

    = 0 N

Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.

Answer:

x = 73.71 [m]

Explanation:

In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.

[tex]v_{f }= v_{i}-(a*t)[/tex]

where:

Vf = fnal velocity = 0

Vi = initial velocity = 32.4 [m/s]

t = time = 4,55 [s]

a = acceleration or desacceleration [m/s^2]

0 = 32.4 - (a*4.55)

a = 7.12 [m/s^2]

Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.

Now using the following equation:

[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]

where:

xo = initial distance = 0

x = final distance [m]

Therefore we have:

x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)

x = 73.71 [m]

Answer:

Explanation:

The frequency of wave is directly proportional to velocity

f = kV

k = f/V

f1/V1 = f2/V2

Given

f1 = 450Hz

V1 = 343m/s

f2 = ?

V2 = 20m/s

Substitute into the formula

450/343 = f2/20

Cross multiply

343f2 = 450×20

343f2 = 9000

f2 = 9000/343

f2 = 26.24Hz

Answer:

The value  is    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

Explanation:

From the question we are told that

  The atom fraction of metal A at point G is [tex] A  =  0.30 \ m[/tex]

   The atom fraction of metal  A at a distance 5000nm from G is  [tex]A_2 = 0.35[/tex]

   The number of atoms per [tex]m^3[/tex] is    [tex]N_h =  9 * 10^{28}[/tex]

    The diffusion coefficient is  [tex]D =   2* 10^{-14 } m^2/s[/tex]

Generally of the concentration of atoms of metal A at G is  

       [tex] N_A = A * N_h [/tex]

=>    [tex] N_A =  0.3  * 9 * 10^{28}[/tex]

=>     [tex] N_A =   2.7 * 10^{28} 2.7 atoms/m^3[/tex]

Generally of the concentration of atoms of metal A at a distance 5000nm from G is  

       [tex]D =  0.35 *9 * 10^{28}[/tex]

=>     [tex]D =  3.15 * 10^{28} \  atoms / m^3[/tex]

The concentration gradient is mathematically represented as

   [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]

=> [tex]\frac{dN_A}{dx}  =  \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]  

=>   [tex]\frac{dN_A}{dx}  = 9 * 10^{20} / m^4[/tex]  

Generally the flux of the atoms per unit  area according to Fick's Law  is mathematically represented as

       [tex]J =  -D* \frac{d N_A}{dx}[/tex]

=>    [tex]J =  -2* 10^{-14 * 9 * 10^{20} [/tex]

=>    [tex] J =  18*10^{6}\   atoms\ crossing\ /m^2 s  [/tex]

Generally if the cross-section area is [tex] a  =  1 cm^2 =  10^{-4} \  m^2[/tex]

Generally the number of atom crossing the above area  per second is mathematically is  

      [tex]H  =  18*10^{6}    *  10^{-4} [/tex]

=>    [tex]H  =  18*10^{2} \  Atom / sec  [/tex]

Answer: The options are not given.

Here are the options.

a) There is an additional force lifting up on you.

(b) At the top you continue going straight and the seat moves out from under you.

(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (

d) Both b and c are correct.

(e) a, b, and c are correct.

The correct option Is D.

B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.

Explanation:

At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.

That is the scenarion that occur...

Answer:

881.84905 LBS

Explanation:

ThErE :p

3922.66 newtons.

This is an exact amount, to get newtons form kg, multiply by 9.8, or in this case, 10.

This gives you 4000 newtons