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Answer:

F = ma - with this you will get 2.6 m/s^2

Explanation:

Use this formula to get the acceleration...

where F is force, M is mass and A is acceleration

by using this we get...

65 = 25 * a

so, a = 65/25

Therefore, the acceleration is 2.6 m/s^2

Hope that helped :)

The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The given data in the problem is;

The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]

Mass of  lawn mower,[tex]\rm m = 25 kg[/tex]

The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]

Acceleration, is found as the ratio of force and the mass.

[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]

Hence, the acceleration of a lawn mower will be 2.6 m/s².

To learn more about acceleration, refer to the link;

https://brainly.com/question/2437624

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Explanation:

(a) The car will leave tangentially to the road, so it will take path d.

(b) Sum of forces on the car in the vertical direction:

∑F = ma

N − mg = 0

N = mg

Sum of forces on the car in the centripetal direction:

∑F = ma

Nμ = m v²/r

mgμ = m v²/r

gμ = v²/r

v = √(grμ)

The maximum velocity is independent of mass, so adding sandbags will not work.

Answer:

A.) 5 mph

B.) 20 mph

Explanation:

Given that you are traveling 3 miles north then 4 miles west before stopping to take a break. If this hike takes 1 hour,

Let's first calculate the displacement be using pythagorean theorem

D = sqrt ( 3^2 + 4^2 )

D = sqrt ( 9 + 16 )

D = sqrt ( 25 )

D = 5

Average velocity = displacement/time

Average velocity = 5/1

Average velocity = 5 mph

Therefore, your average velocity during that time is 5 mph

- You leave your house to go for a drive to clear your head, driving 10 miles over the course of 30 minutes before arriving back home. 

Average velocity = 10/ 0.5

Average velocity = 20 mph

Explanation:

You can solve for volume using radius or diameter.

Sphere Volume   =   4/3 • π • r³     =     ( π •d³)/6

We're given the diameter so let's use that.

Volume = PI * d^3 / 6

Volume = 3.14159 * 3.0^3 / 6

Volume = 3.14159 * 9 / 2

Volume = 14.137 cubic centimeters

Answer:

Explanation: All the spheres interact with other spheres. For example, rain (hydrosphere) falls from clouds in the atmosphere to the lithosphere and forms streams and rivers that provide drinking water for wildlife and humans as well as water for plant growth (biosphere).

Answer:

a)  I = 0.0198 kg m² ,  b)    I = 21.85 kg m²

Explanation:

For this exercise we will use the definition of moment of inertia

        I = ∫ r² dm

For body with high symmetry they are tabulated

sphere  I = 2/5 m r²

bar with respect to  center of mass I = 1/12 m L²

let's calculate the mass of each body

bar

        ρ = m / V

        m = ρ V

        m = ρ l w h

where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm

        m = 32820 1 0.008 0.04

        m = 10.5 kg

Sphere

       M = ρ V

       V = 4/3 pi r³

       M = rgo 4/3 π r³

give us the density 37800 kg / m³ and the radius of 5 cm

       M = 37800 4/3 π 0.05³

       M = 19.8 kg

a) asks us for the moment of inertia of the sphere with respect to its center of mass

        I = 2/5 M r²

        I = 2/5 19.8 0.05²

        I = 0.0198 kg m²

b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes

        I = I_cm + M d2

where d is the distance from the body to the point of interest

        I_cm = 0.0198 kg m²

the distance to the pivot point is

        l = length of the bar + radius of the sphere

        l = 1 + 0.05 = 1.005 m

        I = 0.0198 + 19.8 1.05²

        I = 21.85 kg m²

Answer:

The distance traveled by the second block is 0.197 m

Explanation:

Given;

mass of the small block, m₁ = 0.2 kg

distance traveled by the block, d₁ =0.394 m

time of travel, t = 2 s

mass of the second block, m₂ = 0.4 kg

distance traveled by the second block, d₂ = ?

The work done per unit time on the inclined plane is given by;

[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]

Therefore, the distance traveled by the second block is 0.197 m

Answer:

The mechanical efficiency of the pump is 91.8 %

Explanation:

Given;

input power, p = 44 kw

density of oil, ρ = 860 kg/m³

motor efficiency, η = 90 %

inlet diameter, d₁ = 8 cm

outlet diameter, d₂ = 12 cm

volume flow rate, V = 0.1 m³/s

pressure rise, P = 500kPa

output power  = motor efficiency x input power

output power  = 0.9 x 44 = 39.6 kW

Thus, the mechanical input power = 39.6 kW

The mechanical output power  is given by change in mechanical energy;

[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]

The mechanical efficiency is given by

η = mechanical output power /  mechanical input power

η = 36.347 / 39.6

η = 0.918

η = 91.8 %

Therefore, the mechanical efficiency of the pump is 91.8 %

Answer:

4. 5g

Explanation:

F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.

Answer:

the full segment is:  30 units long

which coincides with answer C in your list

Explanation:

If M is the midpoint, then it divides the segment in two equal parts. Then, we can say that:

RM = MS (equality among the two parts of the divided segment)

replacing RM with "5 x" and MS with "x + 12", we get:

5 x = x + 12

solving for x:

5 x - x = 12

4 x = 12

then x = 3

With this info, we can calculate the length of each half of the segment and consequently its full length:

RM = 5 x = 5 (3) = 15

then the full segment is:  30 units long

Complete question is;

A fisherman’s boat moves up and down periodically due to water waves. The boat travels from its highest to its lowest point, a vertical distance of 0.60 m, in 2.0 s. The horizontal distance between wave crests is 7.0 m. (a) What is the period of the wave motion? (b) What is the frequency of the wave motion? (c) What is the amplitude of each wave

Answer:

A) Period = 4 s

B) Frequency = 0.25 Hz

C) Amplitude = 0.3 m

Explanation:

A) We are told that the distance it takes for the boat to travel from its highest to its lowest point is 2 s.

Thus, t = 2 s

Now, in waves, period (T) it the time between two successive waves.

This means that;

T = 2t

T = 2 × 2

T = 4 s

B) Also, the frequency(f) is given by the formula;

f = 1/T

f = 1/4

f = 0.25 Hz

C) We are given a vertical distance of 0.60 m from highest to lowest point of waves.

Now, amplitude is half of this distance.

Thus;

Amplitude(A) = 0.6/2

A = 0.3 m

Answer:

-15m/s/s

Explanation:

Acceleration = change in speed/ change in time

The change in speed is calculated by subtracting the initial speed from the final speed, so the change in speed is: 50 - 80 = -30m/s. The change in time is 2 - 0 = 2.

So the car acceleration is -30/2 = -15m/s/s

It is negative because it is decelerating.

Hope this helped!

Answer:

a

         [tex]q_1 = -1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = -1.389 *10^{-5} \  C [/tex]

OR

           [tex]q_1 = 1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = 1.389 *10^{-5} \  C [/tex]    

b

[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]

Explanation:

Generally the force exerted on the string is mathematically represented as

[tex]F = k * e[/tex]

substituting values 180 N/m for k and 0.024 m for e

[tex]F = 180 * 0.024[/tex]

[tex]F = 4.32 \ N[/tex]

This force can also equivalent to the electrostatic force between the charges i.e

[tex]F = k * \frac{q^2}{ r^2}[/tex]

substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have

[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]

[tex]q = \sqrt{1.929 *10^{-10}}[/tex]

[tex]q = 1.389 *10^{-5} \ C [/tex]

Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible  algebraic signs  of the charges are

         [tex]q_1 = -1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = -1.389 *10^{-5} \  C [/tex]

OR

[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

D = - 0.5i - 0.25j

Therefore,

Dx = -0.5

Dy = -0.25

Answer:

Static friction prevents a stationary object from moving while kinetic or dynamic friction slows down a moving object.

Explanation:

Static Friction is the maximum force that must be overcome before a stationary object begins to  move, while kinetic or dynamic friction is the maximum force that must be overcome for an object in motion to continue moving at a uniform velocity.

Static friction keeps a stationary object at rest, once the Force of Static friction is overcome, the Force of Kinetic friction is what slows down the moving object.

Answer:

1. a = 9.8 m/s²

2. t = 4.28 s

3. Vf = 42 m/s

4. Vf = 28 m/s  

Explanation:

1.

Since, the body is under free fall motion. Therefore, the value of its acceleration shall be equal to the acceleration due to gravity.

a = 9.8 m/s²

2.

The time taken by the ball to reach the ground can be calculated by using second equation of motion:

h = Vi t + (1/2)gt²

where,

h = height = 90 m

Vi = initial velocity = 0 m/s

t = time taken = ?

Therefore,

90 m = (0 m/s)t + (1/2)(9.8 m/s²)t²

t = √(18.36 s²)

t = 4.28 s

3.

In order to find final velocity we use first equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(4.28 s)

Vf = 42 m/s

4.

when the ball is at height of 50 m, it means it has covered:

h = 90 m - 50 m = 40 m

we use third equation of motion at this point:

2gh = Vf² - Vi²

(2)(9.8 m/s²)(40 m) = Vf² - (0 m/s)²

Vf = √(784 m²/s²)

Vf = 28 m/s

Answer:

steel because being alloy in metal it has free electrons in metals

Answer:

The value is [tex]D = 205.8 \ m [/tex]

Explanation:

The time taken for the column to take a step mathematically represented as

100 steps => 1 minutes => 60 seconds

1 step => t

=> [tex]t = 0.6 \ s [/tex]

Generally the length of the column is mathematically represented as

[tex]D = v * t[/tex]

substituting 343 m/s for v we have

        [tex]D =  343 * 0.6 [/tex]

  =>    [tex]D =  205.8 \  m  [/tex]

Explanation:

1.) Work = Force*Distance

2.) The direction of motion lies in the same direction as the tension in the cable. So the work done by tension would be positive.

3.) The direction of the weight (due to gravity) would act opposite to the direction of motion of the elevator, so work done by gravity becomes negative.

Answer:

 a = √ (a_t² + a_c²)

a_t = dv / dt ,    a_c = v² / r  

Explanation:

In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.

            a = aₓ i ^ + a_y j ^

Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains

         a = √ (a_t² + a_c²)

where the tangential acceleration is

           a_t = dv / dt

the centripetal acceleration is

          a_c = v² / r

Answer:

vector

Explanation:

The position of a particle is vector

Answer:

2.13 m/s^2.

Find the remaining answer in the explanation

Explanation:

Given that a sprinter practicing for the 200-m dash accelerates uniformly from rest at A and reaches a top speed of 41 km/h at the 61-m mark. He then maintains this speed for the next 83 meters before uniformly slowing to a final speed of 34 km/h at the finish line.

1.) The maximum horizontal acceleration will be when she attained her maximum speed. That is,

Speed = 41 km/h

Convert km/h to m/s

(41×1000) / 3600

Speed = 11.39 m/s

Using speed formula to calculate time

Speed = distance/time

11.39 = 61/t

t = 61/11.39

t = 5.36s

Where the distance = 61 m

Maximum acceleration = velocity /time

Maximum acceleration = 11.39/5.36

Maximum acceleration = 2.13 m/s^2

The maximum acceleration value occurs when the sprinter is starting from rest and attaining the maximum speed.

Answer:

Quantitative is the correct term

Answer:

50 degree.

Explanation:

Given that the components of vector A are given as follows: Ax = 5.6 Ay = -4.7

The angle between vector A and B in the positive direction of x-axis will be achieved by using the formula:

Tan Ø = Ay/Ax

Substitute Ay and Ax into the formula above.

Tan Ø = -4.7 / 5.6

Tan Ø = -0.839

Ø = tan^-1(-0. 839)

Ø = - 40 degree

Therefore, the angle between vector A and B positive direction of x-axis will be

90 - 40 = 50 degree.

Answer: 34 km, 21 km  61 degrees north of east

Explanation: distance = 13 + 3 + 18 = 34

   displacement = 13 - 3 = 10

               10^2 + 18^2 = 424

              find the square root of 424 ( 20.5 rounded to 21 )

The total distance walked is 34 km and the resulting displacement is 20.6 km.

a) The total distance is gotten by summing up all the distance.

Total distance = Distance moved east + distance moved north + distance moved west

Total distance = 13 km + 18 km + 3 km = 34 km

b) The displacement is the distance from the beginning point to end point.

Displacement² = 18² + (13 - 3)² = 18² + 10²

Displacement² = 424

Displacement = 20.6 km

Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.

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