If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)

Question 10 options:

There cannot be any forces applied to the ball.


There must be exactly one force applied to the ball.


The net force applied to the ball is zero.


The net force applied to the ball is directed to the right.

Answer:

a = 11.03 [m/s²]

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces is equal to the product of mass by acceleration.

ΣF = m*a

where:

F = force = 160000 [N]

m = mass = 14500 [kg]

a = acceleration [m/s²]

160000 = 14500*a

a = 11.03 [m/s²]

Answer:

C. 4.3 seconds

Explanation:

B 0.56 s is the time period of a twirlers baton.

Centripetal acceleration is defined as the property of the motion of an object which traversing a circular path.

Any object that is moving in a circle and has an acceleration vector pointed towards the center of that circle is known as Centripetal acceleration.

The centripetal acceleration is given by:

a = 4π²R/T²

Given values are:

a = 47.8 m/s²

D = 0.76 m  so , R = 0.76/2 = 0.38m

Using this formula,

47.8*T² = 4π² x0.38

T² = [tex]\frac{4*3.14^2*0.38}{47.8}[/tex]

T = 0.56 s

Therefore,

A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2 which have  time period of 0.56 s.

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Answer:

a. 250kg I think it's the right answer. hope it helps:)

Answer:

C.10

Explanation:

because when you divide 50 divided by 5 = 10

Answer:

When traveling behind other vehicles, there should be at least a four second space between your vehicles. When the car in front of you passes a stationary object, slowly count to yourself. If you pass the object before the allotted time, you should back off. When traveling at night or inclement weather, these times should be doubled.

Don't talk on a cell phone while driving. Phones detract from your ability to concentrate on the road and increase your chance of a collision by nearly 400%. If you must use the phone, pull over to a safe, well-lit parking lot and place your call there. After completing your call you may continue on your way.bey all speed limits and signs.

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

Answer:

[tex]F = 5.4*10^{-8}\ N[/tex]

Explanation:

Given

Represent the mass of the student with M and the mass of the slice of pizza with m

[tex]M = 92kg[/tex]

[tex]m = 550g[/tex]

[tex]d = 25cm[/tex]

Required

Determine the force of attraction

This is calculated as:

[tex]F = \frac{GMm}{d^2}[/tex]

Where G = gravitational constant

[tex]G = 6.67408 * 10^{-11}\ m^3 kg^{-1} s^{-2}[/tex]

Convert both mass to kilogram and distance to metre

[tex]m = 550g[/tex]

[tex]m = 550kg/1000[/tex]

[tex]m = 0.55kg[/tex]

[tex]d = 25cm[/tex]

[tex]d = 25m/100[/tex]

[tex]d = 0.25m[/tex]

Substitute these values in [tex]F = \frac{GMm}{d^2}[/tex]

[tex]F = \frac{6.67408 * 10^{-11} * 92 * 0.55}{0.25^2}[/tex]

[tex]F = \frac{6.67408 * 92 * 0.55* 10^{-11} }{0.25^2}[/tex]

[tex]F = \frac{337.708448* 10^{-11} }{0.0625}[/tex]

[tex]F = 5403.335168* 10^{-11}[/tex]

[tex]F = 5.403335168* 10^3*10^{-11}[/tex]

[tex]F = 5.403335168*10^{3-11}[/tex]

[tex]F = 5.403335168*10^{-8}[/tex]

[tex]F = 5.4*10^{-8}\ N[/tex]

Answer:

(a) the change in length of the silk is 0.001585 cm

(b) the maximum weight that a single thread can support is 17.67 N

Explanation:

Given;

mass of the spider, m = 0.50 g = 0.5 x 10⁻³ kg

length of the silk, L = 12 mm = 0.012 m

diameter of the silk, d = 0.15 mm

radius of the silk, r = d / 2 = 0.075 mm = 0.075 x 10⁻³ m

The cross sectional area of the silk;

A = πr² = π(0.075 x 10⁻³)²

A = 1.767 x 10⁻⁸ m²

The Young's modulus of elasticity of spider-silk is given by;

2.1 Gpa = 2.1 x 10⁹ N/m²

(a)

Apply  Young's modulus of elasticity equation to determine the change in length of the silk;

[tex]E = \frac{FL}{Ax} = \frac{F_gL}{Ax}\\\\x = \frac{F_gL}{AE}\\\\x = \frac{(0.5*10^{-3}*9.8)(0.12)}{(2.1*10^9)(1.767*10^{-8})}\\\\x = 1.585*10^{-5} \ m\\\\x = 0.01585 \ mm[/tex]

[tex]x = 0.001585 \ cm[/tex]

(b)

the maximum weight that a single thread can support is given by;

[tex]T_{tensile \ strength} = \frac{F_{max}}{A}[/tex]

The tensile strength of spider-silk is given by 1 Gpa = 1 x 10⁹ N/m²

[tex]F_{max} = T_{tensile \ strength}*A\\\\F_{max} = (1*10^9)(1.767*10^{-8})\\\\F_{max} = 17.67 \ N[/tex]

Answer:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!  

The ball rotates 6.78 revolutions.

Explanation:

Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!        

At the bottom the ball has the following angular speed:

[tex] \omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s [/tex]

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

[tex] sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m [/tex]

To find the revolutions we need the time, which can be found using the following equation:                

[tex] v_{f} = v_{0} + at [/tex]  

[tex] t = \frac{v_{f} - v_{0}}{a} [/tex] (1)

So first, we need to find the acceleration:

[tex] v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L} [/tex]    (2)  

By entering equation (2) into (1) we have:

[tex] t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}} [/tex]

Since it starts from rest (v₀ = 0):  

[tex] t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s [/tex]

Finally, we can find the revolutions:  

[tex] \theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev [/tex]

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

Answer:

The head-to-tail method is a graphical way to add vectors, The tail of the vector is the starting point of the vector, and the head (or tip) of a vector is the final, pointed end of the arrow.

Explanation:

Answer:

a) The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].

b) The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.

c) The maximum kinetic energy is increased by a factor of 9.

Explanation:

a) From Mechanical Physics, we remember that the mechanical energy of mass-spring system ([tex]E[/tex]), measured in joules, is the sum of the translational kinetic energy ([tex]K[/tex]), measured in joules, and elastic potential energy ([tex]U[/tex]), measured in joules. That is:

[tex]E = K + U[/tex] (1)

By definitions of translational kinetic energy and elastic potential energy, we have the following expressions:

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

[tex]U = \frac{1}{2}\cdot k\cdot x^{2}[/tex] (3)

Where:

[tex]m[/tex] - Mass, measured in kilograms.

[tex]v[/tex] - Velocity of the mass, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]x[/tex] - Elongation of the spring, measured in meters.

If we know that [tex]U = \frac{1}{9}\cdot E[/tex], [tex]k = k[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then:

[tex]\frac{1}{18}\cdot k\cdot A^{2} = \frac{1}{2}\cdot k\cdot x^{2}[/tex]

[tex]\frac{1}{9}\cdot A^{2} = x^{2}[/tex]

[tex]x= \frac{1}{3}\cdot A[/tex]

The potential energy of the system is 1/9 of the total mechanical energy, when [tex]x= \frac{1}{3}\cdot A[/tex].

b) If we know that [tex]k = k[/tex], [tex]x = \frac{1}{2}\cdot A[/tex] and [tex]E = \frac{1}{2}\cdot k \cdot A^{2}[/tex], then the equation of energy conservation associated with the system is:

[tex]\frac{1}{2}\cdot k\cdot A^{2} = \frac{1}{4}\cdot k\cdot A^{2}+K[/tex]

[tex]K = \frac{1}{4}\cdot k\cdot A^{2}[/tex]

The fraction of the total mechanical energy that is kinetic if the displacement is 1/2 the amplitude is 1/2.

c) From the Energy Conservation equation associated with the system, we know that increasing the amplitude by a factor of 3 represents an increase in the elastic potential energy by a factor of 9. Then, the maximum kinetic energy is increased by a factor of 9.

Answer:

The number is  [tex]N = 300[/tex]

Explanation:

From the question we are told that

   The  net charge is  [tex]Q = -4.8 *10^{-17 } \ C[/tex]

Generally the charge on a electron is [tex]e = - 1.60 *10^{-19 } \ C[/tex]

Generally the number of excess electrons is mathematically represented as

      [tex]N = \frac{Q}{e}[/tex]

=>  [tex]N = \frac{-4.8 *10^{-17}}{-1.60 *10^{-19}}[/tex]

=>  [tex]N = 300[/tex]

Answer:

b.dance

Explanation:

don't know the explanation

Answer:

h' = 55.3 m

Explanation:

First, we analyze the horizontal motion of the projectile, to find the time taken by the arrow to reach the orange. Since, air friction is negligible, therefore, the motion shall be uniform:

s = vt

where,

s = horizontal distance between arrow and orange = 60 m

v = initial horizontal speed of the arrow = v₀ Cos θ

θ = launch angle = 30°

v₀ = launch speed = 35 m/s

Therefore,

60 m = (35 m/s)Cos 30° t

t = 60 m/30.31 m/s

t = 1.98 s

Now, we analyze the vertical motion to find the height if arrow at this time. Using second equation of motion:

h = Vi t + (1/2)gt²

where,

Vi = Vertical Component of initial Velocity = v₀ Sin θ = (35 m/s)Sin 30°

Vi = 17.5 m/s

Therefore,

h = (17.5 m/s)(1.98 s) + (1/2)(9.81 m/s²)(1.98 s)²

h = 34.6 m + 19.2 m

h = 53.8 m

since, the arrow initially had a height of y = 1.5 m. Therefore, its final height will be:

h' = h + y

h' = 53.8 m + 1.5 m

h' = 55.3 m

Answer:

B

Explanation:

To solve this we must be knowing each and every concept related to  electrical energy.  Therefore, the correct option is option B among all the given options.        

The work done by an electric charge is referred to as electrical energy. For time t seconds, if electricity I ampere passes throughout a conductor or any other conducting element with a potential differential v volts across it.

The kilowatt hour is both a practical as well as an economic unit of electrical energy. The basic commercial unit is the watt-hour, one and kilowatt hour equals 1000 watt hours. Electric supply providers charge their customers every kilowatt hour unit of electricity used. This kilowatt hour seems to be a BOT unit, or board of trade unit.  The electrical energy that does not become light energy, the lightbulb transforms it into thermal energy.

Therefore, the correct option is option B among all the given options.        

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Answer:

a.  -4.166 J/K

b. 8.37 J/K

c. 4.21 J/K

d. entropy always increases.

Explanation:

Given :

Temperature at hot reservoir , [tex]$T_h$[/tex] = 720 K

Temperature at cold reservoir , [tex]$T_c$[/tex] = 358 K

Transfer of heat, dQ = 3.00 kJ = 3000 J

(a). In the hot reservoir, the change of entropy is given by:

[tex]$dS_h= -\frac{dQ}{t_h}$[/tex]              (the negative sign shows the loss of heat)

[tex]$dS_h= -\frac{3000}{720}$[/tex]

      =  -4.166 J/K

(b)  In the cold reservoir, the change of entropy is given by:

[tex]$dS_c= \frac{dQ}{t_c}$[/tex]              

[tex]$dS_c= \frac{3000}{358}$[/tex]

      =  8.37 J/K

(c). The entropy change in the universe is given by:

[tex]$dS=dS_h+dS_c$[/tex]

    = -4.16+8.37

   = 4.21 J/K

(d). According to the concept of entropy, the entropy of the universe is always increasing and never decreasing for an irreversible process. If the entropy of universe decreases, it violates the laws of thermodynamics. Hence, in part (c), the result have to be positive.

Answer:

A. linear

Explanation:

because they are going in a straight line

Answer:the answer is D

Explanation: I just took the quiz

Answer:

The ratio is  [tex]\frac{B_1}{B_2} = 1.265[/tex]

Explanation:

From the question we are told that

   The density of fresh water is  [tex]\rho__{f}} = 998 \ kg/m^3[/tex]

     The density of ethanol is  [tex]\rho_{e} = 789 \ kg /m^3[/tex]

Generally speed of a wave in a substance is mathematically represented as

    [tex]v = \sqrt{\frac{B}{\rho} }[/tex]

Here B is the adiabatic bulk modulus of the substance  while [tex]\rho[/tex]  is the density of the substance

So at constant wave speed

     [tex]\sqrt{\frac{B_1}{\rho_1} } = \sqrt{\frac{B_2}{\rho_2} }[/tex]

=>   [tex]\frac{B_1}{\rho_1} = \frac{B_2}{\rho_2}[/tex]

=>   [tex]B_1 \rho_2 = B_2\rho_1[/tex]    

=>  [tex]\frac{B_1}{B_2} = \frac{\rho_1}{\rho_2}[/tex]

Here  [tex]\rho_1 =\rho__{f}} = 998 \ kg/m^3[/tex]   and  [tex]\rho_2 = \rho_{e} = 789 \ kg /m^3[/tex]

So

  =>  [tex]\frac{B_1}{B_2} = \frac{998}{789}[/tex]    

  =>  [tex]\frac{B_1}{B_2} = 1.265[/tex]    

Answer:

This is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.

Gravitational field strength = Weight/mass unit is N/kg

Weight = mass x gravitational field strength unit is N

On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.

Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.

Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.

Answer:

However, those astronauts would experience a second spectacle: A solar eclipse caused by the Earth – the Sun disappearing behind the dark disc of the Earth. When Earth inhabitants witness a lunar eclipse, Moon inhabitants would, simultaneously be witnessing a solar eclipse.

Answer:

A) all substance conduct electricity

Answer:

a) The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].

b) The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex]. The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].

c) The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].

d) The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

Explanation:

a) The kinetic energy of the particle is entirely translational, whose formula is:

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

Where:

[tex]K[/tex] - Translational kinetic energy, measured in joules.

[tex]m[/tex] - Mass of the particle, measured in kilograms.

[tex]v[/tex] - Velocity of the particle, measured in meters per second.

The velocity of the particle is the rate of change of the position of the particle in time, that is:

[tex]v = 5+6\cdot t^{2}[/tex] (2)

Where [tex]t[/tex] is the time, measured in seconds.

By substituting on (1), we have the following expression: ([tex]m = 4\,kg[/tex])

[tex]K = 2\cdot (5+6\cdot t^{2})^{2}[/tex]

[tex]K = 2\cdot (25+60\cdot t^{2} +36\cdot t^{4})[/tex]

[tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex]

The kinetic energy of the particle at any time t is [tex]K = 50+120\cdot t^{2}+72\cdot t^{4}[/tex].

b) The acceleration of the particle is the rate of change of the velocity of the particle in time, that is:

[tex]a= 12\cdot t[/tex] (3)

Where [tex]a[/tex] is the acceleration of the particle, measured in meters per square second.

The acceleration of the particle at time t is [tex]a= 12\cdot t[/tex].

The force is obtained by multiplying (3) by the mass of the particle. That is to say: ([tex]m = 4\,kg[/tex])

[tex]F = m\cdot a[/tex] (4)

[tex]F = 48\cdot t[/tex]

The force acting on the particle at time t is [tex]F = 48\cdot t[/tex].

c) According to the Work-Energy Theorem, the change in kinetic energy of the particle equals the change in the net work done on the particle. In this case, the power is equal to the rate of change in kinetic energy.

[tex]\dot W = \dot K[/tex] (5)

[tex]\dot W = \frac{d}{dt}(50+120\cdot t^{2}+72\cdot t^{4})[/tex]

[tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex]

The power being delivered to the particle at time t is [tex]\dot W = 240\cdot t +288\cdot t^{3}[/tex].

d) The work done on the particle ([tex]W[/tex]), measured in joules, is equal to the change of the kinetic energy of the particle:

[tex]W = K(5)-K(0)[/tex] (6)

[tex]W = [50+120\cdot (5)^{2}+72\cdot (5)^{4}]-[50+120\cdot (0)^{2}+72\cdot (0)^{4}][/tex]

[tex]W = 48000\,J[/tex]

The work done on the particle in the interval t = 0 to t = 5 is 48000 joules.

Answer:

Explanation:

horizontal component of velocity of throw = 20 cos20 = 18.8 m /s

vertical downwards component = 20 sin20 = 6.84 m /s

time to displace by height 30 m = t  , initial velocity u = 6.84 m /s

h = ut + 1/2 gt²

30 = 6.84 t + .5 x 9.8 t²

4.9 t² + 6.84 t - 30 = 0

t = - 6.84 ±√( 6.84² + 4 x 4.9 x 30 ) / 2x 4.9

=  - 6.84 ±√( 46.78 + 588 ) / 9.8

=  - 6.84 ±√(634.78 ) / 9.8

= - 6.84 ±25.2 / 9.8

= 1.87 s

horizontal displacement in 1.87 s

= 18.8 x 1.87

= 35.15 m .

Answer:

Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.

Explanation:

Answer:

Two ways to determine ones muscular strength is to either lift the heaviest weight possible or by doing one repetition. The advantages these two methods is that it is easy to determine and does not require calculations or estimating.

Explanation:

the emitter of x-rays rotates around the patient and the detector, placed in diametrically opposite side, picks up the image of a body section (beam and detector move in synchrony)

i believe that should be ur answer =) good luck !

Answer:

The value is  [tex]\alpha = 24.5 \ rad/s^2[/tex]

Explanation:

From the question we are told that

  The weight of the elevator is [tex]W = 7300 \ N[/tex]

  The acceleration it is to be given is  [tex]a = 0.150 \ g = 0.150 * 9.8 = 1.47 \ m/s^2[/tex]

  The diameter of the shaft is [tex]d = 12.0 \ cm = 0.12 \ m[/tex]

Generally the radius is mathematically represented as

      [tex]r = \frac{d}{2}[/tex]

=>   [tex]r = \frac{0.12}{2}[/tex]

=>   [tex]r = 0.06 \ m[/tex]

Generally the minimum angular acceleration is mathematically represented as

       [tex]\alpha = \frac{a}{r}[/tex]

=>    [tex]\alpha = \frac{1.47}{0.06}[/tex]

=>    [tex]\alpha = 24.5 \ rad/s^2[/tex]

Answer:

The correct answer is option B

Explanation:

Step one:

given data

a. force F= 110N

distance s= 3meters

we know that work= Force* distance

work= 110*3

Work= 330Joules

Step two:

data

Force= 60 pounds

distance= 30 ft

convert pounds to Newton

1 pound= 4.44822N

60 pounds= 60*4.44822

=266.9N

convert ft to meteres

1 ft = 0.3048meter

30ft= 0.3048*30

=9.144N

we know that work= Force* distance

work= 266.9N*9.144N

Work= 2440.53Joules

Answer:

Salt melts ice and helps keep water from re-freezing by lowering the freezing point of water. This phenomenon is called freezing point depression. Salt only helps if there is a little bit of liquid water available. The salt has to dissolve into its ions in order to work.

Explanation: