A track star runs a 100m race in 12s what is the velocity of the runner?

Answer:

The distance is 58.03 m

Explanation:

Constant Acceleration Motion

It occurs when the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

[tex]v_f=v_o+at[/tex]

The distance traveled by the object is given by:

[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}[/tex]

The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.

The final speed is:

[tex]v_f=20.7+(-3.4)\cdot 7.8[/tex]

[tex]v_f=20.7-26.52[/tex]

[tex]v_f=-5.82\ m/s[/tex]

Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:

[tex]\displaystyle x=20.7\cdot 7.8+\frac{(-3.4)\cdot 7.8^2}{2}[/tex]

[tex]\displaystyle x=161.46-103.43[/tex]

x=58.03 m

Answer:

f = 3.97 Hz

Explanation:

Given that,

Centripetal acceleration, [tex]a=13\ m/s^2[/tex]

The radius of motion is 0.02 m

The formula for the centripetal acceleration is given by :

[tex]a=\dfrac{v^2}{r}\\\\v=\sqrt{ar} \\\\v=\sqrt{13\times 0.02} \\\\v=0.5\ m/s[/tex]

The speed of an object in a circular path is given by :

[tex]v=\dfrac{2\pi r}{t}[/tex]

t is time period

Also, f=1/t (f is frequency)

[tex]f=\dfrac{v}{2\pi r}\\\\f=\dfrac{0.5}{2\pi \times 0.02}\\\\f=3.97\ Hz[/tex]

Hence, the frequency of motion s 3.97 Hz.

The frequency of the motion is 4.1 Hz.

The linear velocity of the of the object is calculated as follows;

[tex]a = \frac{v^2}{r} \\\\v^2 = ar\\\\v = \sqrt{ar} \\\\v = \sqrt{13 \times 0.02} \\\\v = 0.51 \ m/s[/tex]

The angular speed of the object is calculated as follows;

[tex]\omega =\frac{v}{r} \\\\\omega = \frac{0.51}{0.02} \\\\\omega = 25.5 \ rad/s[/tex]

The frequency of the motion is calculated as follows;

[tex]\omega = 2\pi f\\\\f = \frac{\omega }{2\pi} \\\\f = \frac{25.5}{2\pi } \\\\f = 4.1 \ Hz[/tex]

Thus, the frequency of the motion is 4.1 Hz.

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131.2 J and The last one on number 8

I gave the same answer and it passed.

7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.

8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.

Given:

A moving object is rolling on a surface that is 5 m off the ground,

The speed of the object, v = 4 m/s,

The mass of the object, m = 3.2 kg,

Calculate the kinetic energy after 10 meters as shown below,

KE = 1/2 × 4² × 3.2

KE = 25.6 J

Calculate the potential energy as shown below,

PE = 3.2 × 9.8 × 5

PE = 156.8 J

Thus, total energy = KE + PE

The total energy = 25.6 + 156.8

The total energy = 182.4 J

8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.

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Distance is the total path covered by the object

Here, 200 m is the distance covered by the person and NOT the displacement

Displacement of an object is nothing more than the shortest path between the initial and the final point

If the person travelled 100m and came back, his initial and final point will remain the same which means that he will have a displacement of 0 m

Answer:

c.Mechanical

Explanation:

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

The speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

Given the following data:

To find the speed of the cart at the end of this 3.0 second interval, we would use the first equation of motion;

[tex]V = U + at[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]V = 4 + 0.5(3)\\\\V = 4 + 1.5[/tex]

Final velocity, V = 5.5 m/s.

Therefore, the speed of the cart at the end of this 3.0 second interval is 5.5 meter per seconds.

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Answer:

what's your question I can't understand

Answer:

 m = 876.71 kg

Explanation:

This is an exercise of Archimedes' principle, which states that the thrust on a body is equal to the weight of the dislodged liquid  

        B = ρ g V  

therefore the load that the balloon can lift is  

       B - W_structure - w_load = 0

       w_load = B - W_structure

The volume of the balloon is  

      v = 4/3 π r³

let's substitute  

      w_carga = rho g 4/3 π r³ - m_structure g  

the air density at T = 25ºc is ρ = 1.18 kg / m³

let's calculate  

     w_load = 1.18 9.8 4/3 π 7.15³ - 930 9.8  

     w_load = 17705,77 - 9114  

     w_ load = 8591.77 N

this corresponds to a mass of  

   w_load = m g  

   m = w_load / g  

   m = 8591.77 / 9.8  

   m = 876.71 kg

Answer:

B

Explanation:

74 turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H

The solenoid is a type of electromagnet and the main purpose of the solenoid is to generate a controlled magnetic field through a coil wound into a tightly packed helix.

The solenoid  is a coil structure  of wire, and the plunger is made of soft iron. The magnetic field is formed around it when an electric current passes through it and draws the plunger in.

the solenoid is responsible for converting electrical energy into mechanical work, these were the results of the development of an efficient and greater strength offering magnets which was discovered in the year 1823.

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Answer:

The value is  [tex]w = 0.1167 \ rev/second[/tex]

Explanation:

From the question we are told that

    The rate at which the plate rotates is  [tex]w =7.0 \ rev/min[/tex]

Generally the revolution per second is mathematically represented as

       [tex]w = \frac{7.0}{60}[/tex]

=>    [tex]w = 0.1167 \ rev/second[/tex]

Answer:

v = 17.5 m/s = 63 km/h

Explanation:

       [tex]K = \frac{1}{2}*m *v^{2} (1)[/tex]

       where m = mass of the object, v= speed of the object.

       [tex]21 km/hr * \frac{1 hr}{3600s}*\frac{1000m}{1 km} = 5.83 m/seg (2)[/tex]

       [tex]K = \frac{1}{2}*18000 kg *(5.83m/s)^{2} = 306250 J (3)[/tex]

       [tex]K = 306250 J = \frac{1}{2}*2000 kg *v^{2} (4)[/tex]

       [tex]v = \sqrt{\frac{306250}{1000} (m/s)2} = 17. 5 m/s = 63 km/h (5)[/tex]

Answer:

a

Explanation:

A 60-W light bulb emits spherical electromagnetic waves uniformly in all directions. If 50% of the power input to such a light bulb is emitted as electromagnetic radiation, what is the radiation intensity at a distance of 2.00 m from the light bulb    

so i did 60(50%)=30÷2=15

The radiation intensity is 15 W / m².

To find the radiation intensity, the values are given as,

Power = 60 W

Distance = 2 m

50% of the power input is emitted as electromagnetic radiation.

      The amount of energy emitted per unit solid angle by per unit area of the radiating surface can be said as radiation intensity.

     As, when there is a power output, the input power will emit some energy as a kinetic energy and electromagnetic radiation. By the way, the power input emits 50 % as a radiation, so the power input given as,

                   P = 60 / 2 ( As it was 50 % )

                      = 30 Watt.

The radiation intensity is,

                In = 30 / 2

                    =  15 W / m².

Thus, the radiation intensity is calculated as, 15 W/ m².

So, Option A is the correct answer.

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Answer:

C

Explanation:

Answer:

w = 0.0436 rad/s

Explanation:

Radius of the cylinder is

3.125 * 1609m/mile = 5028 m

Using the angular acceleration formula, we have

a = v²/r = w²r.

Since we're interested in the angular speed, we make w the subject of the formula, so that

w² = a/r

w = √a/r

w = √(9.8/5028)

W = √0.0019049

w = 0.0436 rad/s

Then we can conclude that the angular speed of the rotation is w = 0.0436 rad/s

(a) The ball's height y at time t is given by

y = (20 m/s) sin(40º) t - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :

0 = (20 m/s) sin(40º) t - 1/2 g t ²

0 = t ((20 m/s) sin(40º) - 1/2 g t )

t = 0   or   (20 m/s) sin(40º) - 1/2 g t = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 g t

t = (40 m/s) sin(40º) / g

t ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So

0² - ((20 m/s) sin(40º))² = 2 (-g) y

where y in this equation refers to the maximum height of the ball. Solve for y :

y = ((20 m/s) sin(40º))² / (2g)

y ≈ 8.4 m

Answer:

angle minimum   θ = 41.3º

Explanation:

For this exercise let's use Newton's second law in the condition of static equilibrium

    N - W = 0

    N = W

The rotational equilibrium condition, where we place the axis of rotation on the wall

We assume that counterclockwise rotations are positive

     fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0

the friction force formula is

     fr = μ N

     fr = μ W

we substitute

      μ m g l sin θ - m g l cos θ + mg l /2   cos θ = 0

      μ sin θ - cos θ + ½ cos θ= 0

       μ sin θ - ½ cos θ = 0

       sin θ / cos θ = 1/2 μ

       tan θ = 1/2 μ

       θ = tan⁻¹ (1 / 2μ)

       θ = tan⁻¹ (1 (2 0.57))

      θ = 41.3º

Answer:

[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]

Explanation:

Coulomb's Law

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]

Where:

[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]

Substituting values:

[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]

[tex]q = 1.05\cdot 10^{-5}~c[/tex]

This charge corresponds to a number of electrons given by the elementary charge of the electron:

[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]

Thus, the charge of any of the spheres is:

[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]

[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]

Answer:

a

Explanation:

The energy transformation occurs after a skydiver reaches terminal velocity is follows as;

A. Gravitational potential energy transforms into thermal energy.

B. Gravitational potential energy transforms into kinetic energy.

The skydiver, when he is located at a certain height h above the ground, possesses gravitational potential energy, equal to:

U = mgh

where m is the mass of the skydiver, g is the gravitational acceleration and h is the height above the ground.

The skydiver gravitational potential energy decreases as the altitude decreases and his kinetic energy store increases as his speed increases.

When a skydiver jumps out of a plane, the energy transfers take place as;

The skydiver's kinetic energy store increases as their speed increases and the thermal store of the air and the skydiver increases, as there is friction between the skydiver and the air particles.

In the given situation, both options A and B are correct.

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Answer:

Explanation:

The pressure transmitted in the hydraulic system can be found by using the formula

[tex]p = \frac{f}{a} \\ [/tex]

f is the force

a is the area

From the question we have

[tex]p = \frac{35}{5} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

Answer:

The ratio is  [tex]R_c:R_e = 4 : 1[/tex]

Explanation:

From the question we are told that

   The period of the satellite is  [tex]T_c = 8 \ years[/tex]

Generally the period of earth around the sun is [tex]T_e = 1 \ year[/tex]

Generally from Kepler's third law , which is mathematically represented as

      [tex]\frac{T_c ^2}{T_e^2} = \frac{R_c^3}{R_e^3}[/tex]

Here  [tex]R_c[/tex] is the radius of the orbit which the satellite rotate around the sun

         [tex]R_e[/tex] is the radius of the orbit which the earth rotate around the sun

=>  [tex]\frac{R_c^3}{R_e^3} = [\frac{8}{1} ]^2[/tex]

=>   [tex]\frac{R_c}{R_e} = \sqrt[3]{[\frac{8}{1} ]^2}[/tex]      

=>   [tex]\frac{R_c}{R_e} = \frac{4}{1 }[/tex]      

=>   [tex]R_c:R_e = 4 : 1[/tex]

Answer:

(a) θ = 45° = 0.78 rad

(b) θ = 32.27° = 0.56 rad

(c) θ = 57.27° = 1 rad

Explanation:

When a vector is resolved into its rectangular components, the formula for the direction angle of the vector with positive x-axis is given as:

tan θ = Ay/Ax

θ = tan⁻¹(Ay/Ax)

(a)

Ax = 12 m

Ay = 12 m

θ = tan⁻¹(12 m/ 12 m)

θ = tan⁻¹(1)

θ = 45° = 0.78 rad

(b)

Ax = 19 m

Ay = 12 m

θ = tan⁻¹(12 m/19 m)

θ = tan⁻¹(0.6315)

θ = 32.27° = 0.56 rad

(c)

Ax = 12 m

Ay = 19 m

θ = tan⁻¹(19 m/12 m)

θ = tan⁻¹(1.58333)

θ = 57.27° = 1 rad

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

From the question we have

[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]

We have the final answer as

Hope this helps you

Answer:

Free Body Diagram

Explanation:

Such picture is called the "free body diagram"

Answer:

1.52g

Explanation:

Given parameters:

Force  = 125N

Mass combined  = 82kg

Unknown:

Acceleration of the bicycle  = ?

Solution:

From Newton second law of motion suggests that:

   Force = mass x acceleration

  Acceleration = [tex]\frac{force}{mass}[/tex]  = [tex]\frac{125}{82}[/tex]   = 1.52g

Answer:

The answer that was correct for me was A. 55 N pulling left, and 16 N, 17N p

pulling Right

Explanation:

Answer:

box 1 has larger mass than box 2  (which agrees with the first answer option given.

Explanation:

We need to consider the linear momentum of the boxes immediately before and after they crash.

Recall that momentum is defined as mass times velocity.

So for before the collision, the linear momentum of the system of two boxes is:

m1 * 4km/h  - m2 * 8km/h

with m1 representing mass "1" on the left, and m2 representing mass 2 on the right.

Notice the sign of the linear momentum (one positive (moving towards the right) and the other one negative (moving towards the left)

For after the collision, we have or the linear momentum of the system:

- m1 * 2km/h - m2 * 1km/h

Then, since the linear momentum is conserved in the collision, we make the  initial momentum equal the final and study the mass relationship between m1 and m2:

4 m1 - 8 m2 = - 2 m1 - m2

combining like terms for each mas on one side and another of the equal sign, we get;

4 m1 + 2 m1 = 8 m2 - m2

6 m1 = 7 m2

therefore m1 = (7/6) m2

which (since 7/6 is a number larger than one) tells us that m1 is larger than m2 by a factor of 7/6

Therefore, the first answer option is the correct answer.

Answer:

A. -5.5 N

Explanation:

I can confirm that the answer is A.

Answer:

8

Explanation:

Answer:

t = 5.16 seconds.

Explanation:

The flight time can be found using the following equation:

[tex] y_{f} = y_{0} +v_{0y}*t - \frac{1}{2}gt^{2} [/tex]

Where:

t: is the time

g: is the gravity = 9.81 m/s²

y₀: is the initial height = 0

[tex]y_{f}[/tex]: is the final height = 5.50 m

[tex]v_{0y}[/tex]: is the initial vertical velocity = v*sin(35)

v: is the speed = 46 m/s

[tex] 5.50 = 46*sin(35)*t - \frac{1}{2}9.81*t^{2} [/tex]

By solving the above cuadratic equation we have:

t₁= 0.22 s and t₂= 5.16 s      

We will take the solution equal to 5.16 seconds, since in 0.22 seconds (very short time) the ball is going up and in 5.16 seconds it landed on the green.                        

Therefore, 5.16 seconds have passed since the ball was hit until it landed.

I hope it helps you!                                                                                            

Answer:

t = 10.31 seconds which agrees with answer option A)

Explanation:

Notice that the horizontal velocity of the plane imparted to the package, doesn't affect the vertical motion for which the original (vertical velocity) is zero.

Then the equation for the distance travelled by the package is:

d = (1/2) a t^2

in our case, d = 521 m, and a = g (acceleration of gravity 9.8 m/s^2)

then we can solve for t in the equation:

521 = 9.8/2   t^2

t^2 = 521/4.9

t^2 = 106.32 s^2

t = 10.31 seconds