Suppose that 6 J of work is needed to stretch a spring from its natural length of 26 cm to a length of 39 cm. (a) How much work (in J) is needed to stretch the spring from 30 cm to 35 cm

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

[tex]s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)[/tex]

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

[tex]T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}[/tex]

s₁ = 1.1 m

Hence, the correct option is:

b. 1.1 m

Answer:27

Explanation:

If a roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds, then the acceleration of the rider would be 4 m / s² , therefore the correct answer is option D.

There are three equations of motion given by  Newton,

v = u + at

S = ut + 1/2 × a × t²

v² - u² = 2 × a × s

As given in the problem, A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds.

By using the first equation of the motion,

v = u + at

16 = 4 + 3a

a = 16 -4 / 3

  = 12 / 3

  =  4 m / s²

Thus, the acceleration of the rider would be 4 m / s².

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Answer:

electrons are located around the nucleus of an atom.

Explanation:

Electron configurations describe where electrons are located around the nucleus of an atom. For example, the electron configuration of lithium, 1s²2s¹, tells us that lithium has two electrons in the 1s subshell and one electron in the 2s subshell.

Answer:

the charge of the particle is -2.144 x 10⁻⁵ C.

Explanation:

The force acting on the particle is calculated as;

F = EQ = mg

[tex]Q = \frac{mg}{E}[/tex]

where;

Q is magnitude of the charge of the particle

[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]

since the magnetic field is acting downward, the force must be acting upward in opposite direction.

Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.

Answer:

1. about 1.5 AU

2. about 5 AU

3. about 8 light-years

4. about 100,000 light-years

5. less than 0.01 AU

Explanation:

a. Mars is about 1.5 AU from the Sun.

b. Jupiter is about 5 AU from the Sun.

c. The star Sirius is about 8 light-years from the Sun.

d. The diameter of the Milky Way Galaxy is about 100,000 light-years.

e. The distance from Earth to the Moon is less than 0.01 AU.

Note: AU is an acronym for Astronomical Unit and it is a standard unit by astronomers to illustrate the distance between the planetary bodies found in the solar system.

Answer:

25.3J

Explanation:

Given parameters:

Mass of aluminum  = 3.05g

Initial temperature  = 10.8 °C

Final temperature  = 20 °C

Specific heat  = 0.9J/g °C

Unknown:

Amount of heat needed for the temperature to change  = ?

Solution:

To solve this problem, we use the expression:

       H  = m C Ф  

H is the amount of heat

m is the mass

C is the specific heat capacity

Ф is the change in temperature

     H  = 3.05 x 0.901 x (20 - 10.8) = 25.3J

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

photoelectric effect

Explanation:

The photoelectric effect supports a particle theory of light in that it behaves like an elastic collision (one that conserves mechanical energy) between two particles, the photon of light and the electron of the metal.

Answer:

well you want to add 88.2 and s2.

Explanation:

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

            Work done  = Weight x height  = 700 x 828

       Work done  = 579600J

Answer:

4116J

Explanation:

Given parameters:

Mass of the person  = 70kg

Height of the ladder  = 6m

Unknown:

Work done  = ?

Solution:

The work done by the person climbing is the same as the potential energy.

Work done is the force applied to move a body through a distance;

So;

   Potential energy  = mass x acceleration due to gravity x height

Therefore;

    Potential energy  = 70 x 9.8 x 6   = 4116J

Answer:

When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.

Explanation:

hope this helps!

Answer:

When the circuit is connected properly, the current starts flowing from one end of the battery to the other end.

Explanation:

Edmentum answer.

Answer:

The Law of Conversation

Answer:

kinetic energy

kinetic energy

Answer: take less time to go the same distance

Explanation:

Because if it is going faster let’s say mph 60 mph is 60 miles per hour if you are going 40 miles per hour it will take you longer to get to your destination.

Answer:

displacement is the distance between the starting point and the ending point of an object's journey

Answer:

C. Is traveling on a curve in the road

    Hope this helps :3

The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.

Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.

The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.

Here, all the three vehicles  are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.

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The image related with this question is attached below:

Answer:

Explanation:

Fluid A :

Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³

volume strain = Δ V / V  = 196 x 10⁻⁶ / 3000 x 10⁻⁶

= .06533

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .

It is Acetone .

Fluid B :

Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³

volume strain = Δ V / V  = 138 x 10⁻⁶ / 3000 x 10⁻⁶

= .046

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43  x 10⁷ Pa = 1.3  GPa .

It is Gasoline  .

Fluid C :

Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³

volume strain = Δ V / V  = 84 x 10⁻⁶ / 3000 x 10⁻⁶

= .028

Δ P = increase in pressure = 6 x 10⁷ Pa

E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14  GPa .

It is Water   .

Answer:

static friction

Explanation:

static friction is the friction force that acts on objects at rest

Acceleration = (change in velocity) / (time for the change)

Change in velocity = (ending velocity) - (starting velocity)

Change in the plane's velocity = (10,000 m/s north) - (8,000 m/s north)

Change in the plane's velocity = 2,000 m/s north

Time for the change = 40 seconds

Acceleration = (2,000 m/s north) / (40 seconds)

Acceleration = 50 m/s² north

Answer:

The maximum detection range is 39.75 km

Explanation:

Given that;

Antenna height h1 = 40 m

Target height  ( patrol boat mast ) h2 = 11 m

Using RADAR, what is the maximum detection range = ?

Using RADAR

we know that; Maximum detection range = (√17h1 + √17h2) km

where h1 and h2 are heights of the antenna and target height in meters

so we substitute in our values

Maximum detection range = (√(17 × 40) + √(17 × 11)) km

Maximum detection range = (√680 + √187) km

Maximum detection range = (26.0768 + 13.6747) km

Maximum detection range = (26.0768 + 13.6747) km

Maximum detection range = 39.75 km

Therefore, The maximum detection range is 39.75 km

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm=[tex]1900\times 10^{-9}m[/tex]

Where [tex]1nm=10^{-9} m[/tex]

[tex]\lambda_2=400nm=400\times10^{-9}m[/tex]

[tex]\lambda_1=700nm=700\times 10^{-9}m[/tex]

We have to find  the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

[tex]\theta=sin^{-1}(\frac{m\lambda}{d})[/tex]

Using the formula

m=1

[tex]\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})[/tex]

[tex]\theta=21.62^{\circ}[/tex]

Now, m=2

[tex]\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})[/tex]

[tex]\theta_2=24.90^{\circ}[/tex]

[tex]\Delta \theta=\theta_2-\theta_1[/tex]

[tex]\Delta \theta=24.90-21.62[/tex]

[tex]\Delta \theta=3.28^{\circ}[/tex]

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

Answer:

elapsed time is 10.53 sec

Explanation:

Given that;

velocity of the car [tex]V_{P}[/tex] = 20.0m/s

Initial velocity of the SUV  [tex]U_{suv}[/tex] = 30.0 m/s

SUV slow down at a constant a = 1.90 m/s²

Now, Let Vs  represent the final velocity of the SUV and after time t when the police can cross it and d represent the distance both of the cars covers between the two crossing

so;

d = [tex]V_{P}[/tex] × t

also d = [tex]U_{suv}[/tex] t - 1/2 at²

so

[tex]V_{P}[/tex] × t  =  [tex]U_{suv}[/tex] t - 1/2 at²

we substitute

20 × t  =  30 × t - 1/2 × 1.9 × t²

20t  = 30t - 1/2 × 1.9 × t²

1/2 × 1.9 × t² = 30t - 20t

1/2 × 1/9 × t² = 10t

t = 2 × 10 / 1.9               ( as t ≠ 0 )

t = 20 / 1.9

t = 10.53 sec

Therefore, elapsed time is 10.53 sec

Answer:

the local atmospheric pressure is  93.63 kPa

the mass of the weights is 156.9 kg

Explanation:

Given that;

Initial pressure of gas = 100 kPa

mass of piston = 10 kg and diameter = 14 cm = 0.14 m

g = 9.81 m/s²

Now,

P_gas = P_atm + P_piston

100 = P_atm + P_piston --------- let this equation 1

P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²

P_piston = 98.1 / (π/4×( 0.14 )²)

P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa

now, from equation 1

100 = P_atm + P_piston

we substitute

100 = P_atm + 6.37

P_atm = 100 - 6.37

P_atm = 93.63 kPa

Therefore, the local atmospheric pressure is  93.63 kPa

Now for pressure of the gas in the cylinder ⇒ 2×initial pressure

Pgas_2 = 2 × 100 = 200 kPa

Pgas_2 = P_atm + P_piston + P_weight

Pgas_2 =  P_gas  + P_weight

we substitute

200 kPa =  100 kPa  + P_weight

P_weight =  200 kPa -  100 kPa

P_weight = 100 kPa =  100,000 Pa

Also;

P_weight = M×g / A

100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)

100,000 × 0.01539 = M × 9.81

1539 = M × 9.81

M = 1539 / 9.81

M = 156.9 kg

Therefore, the mass of the weights is 156.9 kg

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm

Answer:

The roof and walls of the auditorium or cinema hall are generally covered with sound absorbent materials like draperies or compressed fibreboard to reduce reverberation. These materials reduce the formation of echoes by absorbing sound waves.

Explanation:

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Answer:

μk = 0.3

Explanation:

        [tex]f_{frk} = \mu_{k} * F_{n} (1)[/tex]

      [tex]F_{n} = m*g (2)[/tex]

       [tex]W_{ffr} = -F_{fr} * d = - \mu_{k}* m*g*d (3)[/tex]

        [tex]\Delta K = K_{f} - K_{o} (4)[/tex]

       [tex]\Delta K = 0 - \frac{1}{2} *m * v_{o} ^{2} = - \frac{1}{2} *m * v_{o} ^{2} (5)[/tex]

       [tex]W_{ffr} = \mu_{k}* g*d = \frac{1}{2} * v_{o} ^{2} (6)[/tex]

       [tex]\mu_{k} = \frac{1}{2} * v_{o} ^{2} *\frac{1}{g*d} =\frac{(8.0m/s)^{2}}{2*9.8m/s2*11m} = 0.3 (7)[/tex]