7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J

8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?

The sum of the potential energy and the kinetic energy is always constant.

The sum of the potential energy and the kinetic energy is always 0.

The potential energy is always greater than the kinetic energy.

The kinetic energy is always equal to the potential energy.

Answer:

The modal verb must is used to express obligation and necessity. The phrase have to doesn't look like a modal verb, but it performs the same function.

Answer:Highest frequency  =618.89Hz

Lowest frequency=582.22Hz

Explanation:

 The linear velocity of a sound generator  is related to angular velocity and is given as

Vs = rω  where

r = the radius of circular path = 1.0 m

ω is the angular velocity of the sound generator. = 100 rpm

1 rev/min = 0.10472 rad/s

100rpm =10.472 rad/ s

Vs = rω

= 1m x 10.472rad/ s=  10.472m/s

A) Highest frequency  heard by a student in the classroom = Maximum frequency. Using the Doppler effect formulae,

f max = (v/ v-vs) fs

Where , v is the speed of the sound in air at 20 degrees celcius =

343 metres per second

vs is the linear velocity of the sound generator=10.472m/s

fs is the frequency of the sound generator= 600 Hz

f max = (343/ 343 - 10.472) x 600

=343/332.528) x600

=618.89Hz

B) Lowest frequency  heard by a student in the classroom = Minimum frequency

f min = (v/ v+vs) fs

(343/ 343 + 10.472) x 600

=343/353.472) x 600

=582.22hz

Answer:

H' = H/4

Explanation:

By applying the law of conservation of energy to this problem, we know that:

Elastic Potential Energy Stored by Spring = Gravitational Potential Energy of ball

(1/2)kx² = mgH

H = (1/2)kx²/mg   -------------- equation (1)

where,

H = Height reached by the ball

x = compression of spring

k = stiffness of spring

m = mass of ball

g = acceleration due to gravity

Now, if we make the compression to half of its value:

x' = x/2

then:

H' = (1/2)k(x/2)²/mg

H' = (1/4)(1/2)kx²/mg

using equation (1), we get:

H' = H/4

Answer:

Explanation:

Step one:

given data

initial velocity u= 0m/s since the ball is at rest

time of contact t= 0.3s

final velocity v=15m/s

Required

acceleration a

from the first law of motion

v=u+at

substitute our given data

15=0+a*0.3

15=0.3a

divide both sides by 0.3

a=15/0.3

a=50m/s

The average acceleration is 50m/s^2

Answer:

a

[tex]\lambda = 3.68 *10^{-36} \ m[/tex]

b

[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Explanation:

From the question we are told that

   The mass of the person is  [tex]m = 180 \ kg[/tex]

    The speed of the person is  [tex]v = 1 \ m/s[/tex]

    The energy of the proton is  [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]

Generally the de Broglie wavelength is mathematically represented as

      [tex]\lambda = \frac{h}{m * v }[/tex]

Here  h is the Planck constant with the value

      [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]

So  

     [tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]

=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]

Generally the energy of the proton is mathematically represented as

         [tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]

Here [tex]m_p[/tex]  is the mass of proton with value  [tex]m_p = 1.67 *10^{-27} \ kg[/tex]

=>     [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]

=>   [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]

=>   [tex]v = 3.09529 *10^{7} \ m/s[/tex]

So

        [tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]

so    [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]

=>     [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]

Answer:

box 1 has larger mass than box 2

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, answer 1 is the correct answer.

Answer:

410,000 kg

Explanation:

Use Newton's second law

F = ma

m = F/a

m = 59,400 N/(.145 m/s) = 410,000 kg

Answer:

B

Explanation:

Answer:

Pressure

Explanation: I took a test

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

[tex]Range=v_i^2\frac{sin(2\theta)}{g}[/tex]

which for our case renders:

[tex]Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m[/tex]

By using the range formula in the motion of a projectile, the ball will land 22.6 meters from where it was hit.

Given that a golfer hits a ball at 15 m/s at an angle of 40 degrees above the horizontal, To calculate how far the ball travelled, we will use the range formula to calculate the total distance travelled.

Range R = [tex]u^{2}[/tex]sin2∅ /g

Where

u = 15 m/s

g = 9.8 m/[tex]s^{2}[/tex]

∅ = 40 degrees

Substitute all the parameters into the above formula.

R = [tex]15^{2}[/tex]sin(2 x 40) / 9.8

R = 225sin80/9.8

R = 221.58/9.8

R = 22.6 m

Therefore, the ball will land 22.6 meters from where it was hit.

Learn more here: https://brainly.com/question/25299583

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.        

To know more about electrical energy, here:

https://brainly.com/question/1698090

#SPJ2

Answer:

Satellites may move north to south, or south to north, or west to east, but never from east to west. When satellites are launched, they always head eastward to take advantage of the Earth's rotation, going more than 1,000 miles per hour near the equator. This saves a lot of fuel.

Answer:

true

Explanation:

Im in k12 and I got an 100%

Answer:

The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?

f= Force (of gravity)=500N

g=acceleration of gravity=1.6m/s^2

m=mass=312kg

m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg

his mass is 312kg

Answer:

B. 1275 kg*m/s

Explanation:

I = F(deltaT) = (deltaP) = mv2- mv1

Therefore,

I = mv2-mv1

m = 850 kg

v2 = 5 m/s

v1 = 3.5 m/s

I = (850)(5)-(850)(3.5)

I = 1275 kg* m/s

Answer:

14 m/s

Explanation:

The following data were obtained from the question:

Mass = 50 kg

Initial velocity (u) = 0 m/s

Height (h) = 10 m

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

Final velocity (v) =?

The velocity (v) with which the person hit the water can be obtained as shown below:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 10)

v² = 0 + 196

v² = 196

Take the square root of both side

v = √196

v = 14 m/s

Therefore, he will hit the water with a speed of 14 m/s

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

  Using the dimensions;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]

  = 268.22m/s

Answer:

ok tbh i didnt cry when neji died but now when i think about it i just cry thinking about how himawari could have met her uncle

Explanation:

Answer:

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion.

Explanation:

please mark me brainlist

Mercury is the only one in liquid state at room temperature. It's used in thermometers because it has high coefficient of expansion. they still use mercury even though it is the poorest conductor of heat.

Answer:

B 0.4 have a nice day and hiiii

Answer:

0.546 [tex]\hat k[/tex]

Explanation:

From the given information:

The force on a given current-carrying conductor is:

[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]

where the length usually in negative (x) direction can be computed as

[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]

[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]

[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]

[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]

where;

current I = 7.0 A

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]

[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]

F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]

F = 0.546 [tex]\hat k[/tex]

Answer:

answer is 61.8 m/s; 32.9

l am not sure

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:

(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]