A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

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

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Answer:

We can help to keep it magnificent for ourselves, our children and grandchildren, and other living things besides us.

Explanation:

5 ways our governments can confront climate change

PROTECT AND RESTORE KEY ECOSYSTEMS

SUPPORT SMALL AGRICULTURAL PRODUCERS

PROMOTE GREEN ENERGY

COMBAT SHORT-LIVED CLIMATE POLLUTANTS

BET ON ADAPTATION, NOT JUST MITIGATION

Answer:

The force is Inertia

Explanation:

This is an example: He turns the ketchup bottle at an angle toward his plate and smacks the bottom of the bottle until the ketchup comes out. How does inertia affect the ketchup in the bottle? The Inertia keeps the ketchup in the bottle. The Jacksons are driving to the lake when a car in front of theirs slams on its brakes.

Your welcome! :)

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

[tex]v = \sqrt{2as}[/tex]

[tex]v = \sqrt{2*18.9*391}[/tex]

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

[tex]V= \sqrt{v^2 + 2aS'}[/tex]

[tex]V = \sqrt{121.57^2+ 2 (-9.92)(332)}[/tex]

V = 90.51 m/s

Answer:

0.9 kg

Explanation:

We would use the equation F=m*a to solve this equation. First, we would need to get mass by itself therefore we divide out acceleration from both sides ( F/a=m*a / a ) acceleration would cancel out and the end equation should look like this ( F/a = m or m = F/a)  After we do that we plug in the numbers 1.35 N / 1.5 m/s^2 we get 0.9 kg, assuming you are using kg.

Answer: 0.90 kilograms

Explanation:

Answer:

A. 1 m/s²

B. 312.5 m

C. 12.5 m/s

Explanation:

We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:

Velocity (Km/h) = 90 Km/h

Velocity (m/s) =?

Velocity (m/s) = Velocity (Km/h) × 1000 / 3600

Velocity (m/s) = 90 × 1000 / 3600

Velocity (m/s) = 90000 / 3600

Velocity (m/s) = 25 m/s

A. Determination of the acceleration.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Time (t) = 25 s

Acceleration (a) =?

v = u + at

25 = 0 + (a × 25)

25 = 0 + 25a

25 = 25a

Divide both side by 25

a = 25/25

a = 1 m/s²

B. Determination of the distance travelled.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Acceleration (a) = 1 m/s²

Distance travelled (s) =?

v² = u² + 2as

25² = 0 + (2 × 1 × s)

625 = 0 + 2s

625 = 2s

Divide both side by 2

s = 625 / 2

s = 312.5 m

C. Determination of the average velocity.

Total distance travelled = 312.5 m

Total time = 25 s

Average velocity =?

Average velocity = Total distance / total time

Average velocity = 312.5 / 25

Average velocity = 12.5 m/s

Answer:

No of proton is 13 and nucleus is 13

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

          ΔA = 2α A ΔT

the area is

         A = π r² = π D² / 4

we substitute

         ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

    α = 1.7 10⁻⁵ ºC⁻¹

we calculate

          ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

          ΔA = 6.49 10⁻⁷ m²

by definition

           ΔA = A_f- A₀

           A_f = ΔA + A₀

           A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

           A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

           A_f = 2,551 10⁻⁴ m²

the area is

           A_f = π D_f² / 4

           A_f = [tex]\sqrt{4 A_f /\pi }[/tex]

           D_f = [tex]\sqrt{4 \ 2.551 10^{-4} /\pi }[/tex]

           D_f = 1.80229 10⁻² m

the change in diameter is

           ΔD = D_f - D₀

           ΔD = (1.80229 - 1.8) 10⁻² m

           ΔD = 0.00229 10⁻² m

           ΔD = 2.29 10⁻⁵ m

Answer: 81.619 kJ

Explanation:

Given

Mass of roller coaster is [tex]m=230\ kg[/tex]

It reaches the steepest hill with speed of [tex]u=6.2\ km/h\ or \ 1.72\ m/s[/tex]

Hill to bottom is 51 m long with inclination of [tex]45^{\circ}[/tex]

Height of the hill is [tex]h=51\sin 45^{\circ}=36.06\ m[/tex]

Conserving energy to get kinetic energy at bottom

Energy at top=Energy at bottom

[tex]\Rightarrow K_t+U_t=K_b+U_b\\\Rightarrow \dfrac{1}{2}mu^2+mgh=K_b+0\\\\\Rightarrow K_b=0.5\times 230\times 1.72^2+230\times 9.8\times 36.06\\\Rightarrow K_b=340.216+81,279.24\\\Rightarrow K_b=81,619.456\ J\\\Rightarrow K_b=81.619\ kJ[/tex]

Answer:

The slope of the position time graph gives the velocity.

Explanation:

The slope of the position time graph gives the value of velocity.

In first graph,

The slope is constant in both the parts but positive . So the velocity is also constant and positive for both the parts.  and more than the second part, so the initial velocity is more than the final velocity.

In the second graph,

The slope is constant in both the parts but negative. So, the velocity is constant but negative for both the parts. Initial velocity is more negative than the final velocity.

Answer:

The correct option is D.

Explanation:

The wavelength of the photon can be calculated with the following equation:

[tex] E = h\frac{c}{\lambda} [/tex]

Where:

E: is the energy of the photon = 2.90 eV

h: is the Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3x10⁸ m/s

λ: is the wavelength

Hence, the photon's wavelength is:

[tex] \lambda = \frac{hc}{E} = \frac{6.62 \cdot 10^{-34} J*s*3.0 \cdot 10^{8} m/s}{2. 90 eV*\frac{1.6 \cdot 10^{-19} J}{1 eV}} = 428 nm [/tex]

Therefore, the correct option is D.

I hope it helps you!

Answer:

the magnitude of momentum is √2≈ b

Explanation:

hope that helped

Answer:

Z = 138.5 Ω

Explanation:

In a series RLC circuit the impedance is

          Z = [tex]\sqrt{R^2 + ( X_L - X_C)^2 }[/tex]

the capacitive impedance is

         X_C = 1 / wC

the inductive impedance is

         X_L = wL

in this exercise indicate that C = 50 10⁻³ F, L = 0.3 H and the frequency is f=60 Hz

angular velocity and frequency are related

         w = 2π f

         w = 2π 60

         w = 376.99 rad / s

let's calculate

        Z = [tex]\sqrt{80^2 + ( 376.99 \ 0.3 - \frac{1}{376.99 \ 50 \ 10^{-3}} )^2 }[/tex]

        Z = [tex]\sqrt{6400 + ( 113.1 - 0.053)^2}[/tex]

        Z = √19179.6

        Z = 138.5 Ω

Answer:

125.6 cm

Explanation:

Applying,

S = 2πr................... Equation 1

Where S = distance moved by the tip in one revolution, r = radius of the rotating fan, π = pie

From the question,

Given: r = 20 cm,

Constant: π = 3.14

Substitute these values into equation 1

S = 2(3.14)(20)

S = 125.6 cm

Hence the distance moved by the tip in one revolution is 125.6 cm

Answer:

a)  [tex]k=19.6N/m[/tex]

b)  [tex]V_m=0.81m/s[/tex]

c)  [tex]a_m=6.561m/s^2[/tex]

d)  [tex]K.E=0.096J[/tex]

e)  [tex]T=0.78sec[/tex] &[tex]F=1.29sec[/tex]

f)   [tex]mx'' + kx' =0[/tex]

Explanation:

From the question we are told that:

Stretch Length [tex]L=0.150m[/tex]

Mass [tex]m=0.30kg[/tex]

Total stretch length[tex]L_t=0.150+0.100=>0.25[/tex]

a)

Generally the equation for Force F on the spring is mathematically given by

[tex]F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}[/tex]

[tex]k=19.6N/m[/tex]

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

[tex]V_m=A\omega[/tex]

Where

A=Amplitude

[tex]A=0.100m[/tex]

And

[tex]\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s[/tex]

Therefore

[tex]V_m=A\omega\\\\V_m=8.1*0.1[/tex]

[tex]V_m=0.81m/s[/tex]

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

[tex]a_m=\omega^2A[/tex]

[tex]a_m=8.1^2*0.1[/tex]

[tex]a_m=6.561m/s^2[/tex]

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

[tex]K.E=\frac{1}{2}mv^2[/tex]

[tex]K.E=\frac{1}{2}*0.3*0.8^2[/tex]

[tex]K.E=0.096J[/tex]

e)

Generally the equation for  the period T is mathematically given by

[tex]\omega=\frac{2\pi}{T}[/tex]

[tex]T=\frac{2*3.142}{8.1}[/tex]

[tex]T=0.78sec[/tex]

Generally the equation for  the Frequency is mathematically given by

[tex]F=\frac{1}{T}[/tex]

[tex]F=1.29sec[/tex]

f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

[tex]mx'' + kx' =0[/tex]

Where

'= signify order of differentiation

Answer:

The process of solving a circular motion problem is much like any other problem in physics class. The process involves a careful reading of the problem, the identification of the known and required information in variable form, the selection of the relevant equation(s), substitution of known values into the equation, and finally algebraic manipulation of the equation to determine the answer. Consider the application of this process to the following two circular motion problems.

Sample Problem #1

A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 900 kg

v = 10.0 m/s

R = 25.0 m

Requested Information:

a = ????

Fnet = ????

To determine the acceleration of the car, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (10.0 m/s)2 / (25.0 m)

a = (100 m2/s2) / (25.0 m)

a = 4 m/s2

To determine the net force acting upon the car, use the equation Fnet = m•a. The solution is as follows.

Fnet = m • a

Fnet = (900 kg) • (4 m/s2)

Fnet = 3600 N

Sample Problem #2

A 95-kg halfback makes a turn on the football field. The halfback sweeps out a path that is a portion of a circle with a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting upon the halfback.

The solution of this problem begins with the identification of the known and requested information.

Known Information:

m = 95.0 kg

R = 12.0 m

Traveled 1/4-th of the circumference in 2.1 s

Requested Information:

v = ????

a = ????

Fnet = ????

To determine the speed of the halfback, use the equation v = d / t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows:

v = d / t

v = (0.25 • 2 • pi • R) / t

v = (0.25 • 2 • 3.14 • 12.0 m) / (2.1 s)

v = 8.97 m/s

To determine the acceleration of the halfback, use the equation a = v2 / R. The solution is as follows:

a = v2 / R

a = (8.97 m/s)2 / (12.0 m)

a = (80.5 m2/s2) / (12.0 m)

a = 6.71 m/s2

To determine the net force acting upon the halfback, use the equation Fnet = m•a. The solution is as follows.

Fnet = m*a

Fnet = (95.0 kg)*(6.71 m/s2)

Fnet = 637 N

In Lesson 2 of this unit, circular motion principles and the above mathematical equations will be combined to explain and analyze a variety of real-world motion scenarios including amusement park rides and circular-type motions in athletics.

Answer:

A ball resting on the edge of a cliff

Explanation:

An object must be in motion or be moving to have kinetic energy. Since the ball is resting on the edge of a cliff, it is not actually moving so it does not have kinetic energy but the resting place of the ball is potential energy.

Answer: hello your question is poorly written attached below is the complete question

answer :

TA = 1.6*10^-24 * 60 * 2,  TB = 1.6*10^-24 * ( 60 + 30 ) * 2  -- ( option 1 )

Explanation:

a = 2m/s^2

Ta = m₁ a = 60 * 1.6 * 10^-24 * 2 ц

Tb - Ta = m₂ a

∴ Tb = m₂ a  + Ta

       = ( 30 * 1.6 * 10^-24 * 2 ) +  ( 60 * 1.6 * 10^-24 * 2 )

= ( 30 + 60 ) * 1.6 * 10^-24 * 2 ц

Answer:

The magnitude of the electric field strength is 6.4 x 10⁵ N/C, directed from positive particle to negative particle.

Explanation:

Given;

charge of each particle, Q = 2 x 10⁻⁷ C

distance between the two charges, r = 15 cm = 0.15 m

distance midway between the charges = 0.075 m

The magnitude of the electric field is calculated as;

[tex]E_{net} = E_{+q} + E_{-q}\\\\E_{net} = \frac{kQ}{r_{1/2}^2} + \frac{kQ}{r_{1/2}^2}\\\\E_{net} = 2(\frac{kQ}{r_{1/2}^2})\\\\E_{net} = 2 (\frac{9\times 10^9 \ \times 2\times 10^{-7}}{0.075^2} )\\\\E_{net} = 6.4\times 10^5 \ N/C[/tex]

The direction of the electric field is from positive particle to negative particle.

Answer:

dislike of or prejudice against people from other countries.

Explanation:

Answer: dislike of or prejudice against people from other countries.

Explanation:

Answer: Electrons move around the nucleus in fixed orbits of equal levels of energy

Explanation:

The statement that accurately represents the arrangement of electrons in Bohr’s atomic model is that the electrons move around the nucleus in fixed orbits of equal levels of energy.

It should be noted that the electrons have a fixed energy level when they travel around the nucleus in with energies which varies for different levels.

Higher energy levels are depicted by the orbits that are far from the nucleus. There's emission of light when the electrons then return back to a lower energy level.

Answer:

Explanation:

Hope this helps... pls vote as brainliest

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]

[tex]v_1t= 0.35[/tex]  ...........(i)

[tex]v_0t= 1.09[/tex]  ...........(ii)

Dividing the equation (ii) by (i)

[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]

∴ [tex]v_0=3.11 \ v_1[/tex]

Now loss of energy  = change in the kinetic energy

[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]

[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]

[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]

If f is average friction force, then

(f)(L) = W

(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

(f)  = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

The Average force of friction is ( F )  = 7.307 * 10⁻³ v₀²

Given data:

Predicted range ( v₁t ) = 0.3503 m

Actual range ( v₀t ) = 1.09 m

mass = 16.3 g

First step : Determine the value of  V₀

[tex]t = \sqrt{\frac{2H}{g} }[/tex]    ,    v₁t  =  0.3503 ,    ( v₀t ) = 1.09 m

To obtain the value of  V₀  

Divide ( v₀t ) by ( v₁t )  =  1.09 / 0.3503 = 3.11 v₁

∴ V₀ = 3.11 v₁

Next step : Determine the average force of friction ( f )

given that loss of energy results in a change in kinetic energy

W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]

    = 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]

∴ W = 7.307 * 10⁻³ v₀²

Average force of friction = W / Actual length

                                         = 7.307 * 10⁻³ v₀² / 1  

∴ Average force of friction ( F )  = 7.307 * 10⁻³ v₀²

Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²

Learn more about average force of friction : https://brainly.com/question/16207943

Your question has some missing data below are the missing data related to your question

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Answer

I hope this help....

Explanation:

Answer:

Hope this helps

Explanation:

Answer: The magnification of the magnifying glass is -2.9

Explanation:

The equation for power is given as:

[tex]P=\frac{1}{f}[/tex]

where,

P = Power = 10 D

f = focal length

Putting values in above equation, we get:

[tex]f=\frac{1}{10}=0.1m=10 cm[/tex]            (Conversion factor: 1 m = 100 cm)

The equation for lens formula follows:

[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]

where,

v = image distance

u = Object distance = 6.55 cm

Putting values in above equation, we get:

[tex]\frac{1}{v}=\frac{1}{10}-\frac{1}{(6.55)}\\\\\frac{1}{v}=\frac{6.55-10}{10\times 6.55}\\\\v=\frac{65.5}{-3.45}=-18.98cm[/tex]

Magnification (m) can be written as:

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

Putting values in above equation, we get:

[tex]m=\frac{-18.98}{6.55}\\\\m=-2.9[/tex]

Hence, the magnification of the magnifying glass is -2.9

Answer:

the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

Explanation:

Given the data in the question;

Velocity of Bonzo [tex]V_{Bonzo[/tex] = 1.1 m/s

Velocity of Ender [tex]V_{Ender[/tex] = -4.3 m/s

the ratio Bonzo/mEnder of the masses of these two enemies = ?

Now, using the law of conservation of momentum.

momentum of both Bonzo and Ender are conserved

so

Initial momentum = final momentum

we have

0 = [tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] + [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]

[tex]m_{Bonzo[/tex] × [tex]V_{Bonzo[/tex] = -[ [tex]m_{Ender[/tex] × [tex]V_{Ender[/tex]  ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ [tex]V_{Ender[/tex] / [tex]V_{Bonzo[/tex] ]

we substitute

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -4.3 m/s / 1.1 m/s ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = -[ -3.9090 ]

[tex]m_{Bonzo[/tex] / [tex]m_{Ender[/tex]  = 3.91

Therefore, the ratio Bonzo/mEnder of the masses of these two enemies is 3.91

Answer:

D because of POE

Explanation:

the reaction force is the ball exerting the same newtons of force back to your leg opposite of the ball, but the reaction force and the action force is never stronger than each-other. and action is only being done on the soccer ball, so process of elimination, the answer is D

Answer:

B

Explanation:

it can only have a different motion if it is acted upon a different ball