A clock mechanism at rest has a period of exactly 1 second. When this clock is observed while moving with a speed of 0.80 c, what is its period?

Answer:

I think its A.........

Explanation:

We need to find the value of following expression :

[tex]\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.38 \times 10^6)^2}[/tex]

Firstly, solving the numerator of the above expression :

[tex]=\dfrac{39.8199\times 10^{-11+24}}{40.7044\times 10^{12}}\\\\=\dfrac{39.8199\times 10^{13}}{40.7044\times 10^{12}}\\\\=9.7827[/tex]

Rounding off the result = 9.78

In scientific notation : [tex]9.78\times 10^0[/tex]

The value of a = 9.78 and k = 0.

Answer:

The  force is  [tex]F =  1164.6\  lbf[/tex]

The time is   [tex]\Delta t =  2.44 \  s[/tex]

Explanation:

From the question we are told that

  The  mass of the car is  [tex]m  =  2500 \ lbm[/tex]

   The  initial velocity of the car is [tex]u  =  25 \ mi/hr[/tex]

   The final  velocity of the car is  [tex]v  =  50 \  mi/hr[/tex]

  The acceleration is  [tex]a =  15 ft/s^2 =  \frac{15 *  3600^2}{ 5280} =  36818.2 \  mi/h^2[/tex]

Generally the acceleration is mathematically represented as

      [tex]a =  \frac{v-u}{\Delta t}[/tex]

=>   [tex]36818.2 =  \frac{50 - 25 }{ \Delta t}[/tex]

=>   [tex]t = 0.000679 \  hr[/tex]

converting to seconds

       [tex]\Delta t =  0.0000679 *  3600[/tex]

=>     [tex]\Delta t =  2.44 \  s[/tex]

Generally the force is mathematically represented as

        [tex]F  =  m * a[/tex]

=>      [tex]F  =  2500 *  15[/tex]

=>      [tex]F  =  37500 \ \frac{lbm *  ft}{s^2}[/tex]

Now converting to foot-pound-second we have  

         [tex]F =  \frac{37500}{32.2}[/tex]

=>        [tex]F =  1164.6\  lbf[/tex]

Answer:

v=37.4 m/s

Explanation:

It is given that,

Mass of a ball, m = 3.7 kg

Initial velocity of the ball is u = 8 m/s

We need to find its velocity after 3 seconds. It is moving downwards. The equation of motion is this case is

v=u+gt

[tex]v=8+9.8\times 3\\\\v=37.4\ m/s[/tex]

So, the velocity of the ball after 3 seconds is 37.4 m/s.

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

[tex]T=\dfrac{38}{8}\\\\T=4.75\ s[/tex]

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

[tex]T=2\pi \sqrt{\dfrac{m}{k}}[/tex]

k = spring stiffness constant

So,

[tex]k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m[/tex]

When the cord is in air,

mg=kx

x = the extension in the cord

[tex]x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m[/tex]

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

The spring stiffness constant is 116.7 N/m and the the unstretched length of the bungee cord is 19.54 m.

The given parameters;

The period of oscillation of the bungee jumper is calculated as follows;

[tex]T = \frac{t}{n} \\\\T = \frac{38}{8} \\\\T = 4.75 \ s[/tex]

The spring stiffness constant is calculated as follows;

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\\sqrt{\frac{m}{k} } = \frac{T}{2\pi} \\\\k = m \times \frac{T^2}{4\pi^2} \\\\k = 65 \times \frac{(4.75)^2}{4\pi ^2} \\\\k = 116.7 \ N/m[/tex]

The extension of the cord is calculated as follows;

[tex]F = kx\\\\mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{65 \times 9.8}{116.7} \\\\x = 5.46 \ m[/tex]

The unstretched length of the bungee cord is calculated as;

[tex]\Delta x = l_2-l_1\\\\l_1 = l_2 - \Delta x\\\\l_1 = 25 - 5.46\\\\l_1 = 19.54 \ m[/tex]

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Answer: Hydrogen Peroxide

Explanation: Hydrogen peroxide breaks down into oxygen and water. As a small amount of hydrogen peroxide generates a large volume of oxygen, the oxygen quickly pushes out of the container. The soapy water traps the oxygen, creating bubbles, and turns into foam.

Answer:

The force between the charges are not affected.

Explanation:

Given;

distance between two equal charges, R = 2m

The force between the charges is given by;

[tex]F = \frac{kq^2}{R^2}\\\\F_1 = \frac{kq_1^2}{R_1^2}\\\\When\ the \ charges \ and \ the \ distance \ are \ divided \ by \ two \ (q_2 = \frac{q_1}{2}, \ R_2 = \frac{R_1}{2} )\\\\ F_2 = \frac{kq_2^2}{R_2^2}\\\\F_2 = \frac{k(q_1/2)^2}{(R_1/2)^2}\\\\F_2= \frac{4k*q_1^2}{4*R_1^2}\\\\F_2 = \frac{k*q_1^2}{R_1^2}\\\\F_2 = F_1[/tex]

Therefore, the force between the charges are not affected.

Explanation:

binary number is a number expressed in the base-2 numeral system or binary numeral system, which uses only two symbols: typically "0" and "1".

Answer:

0.68 m

Explanation:

Since the volume of water is 310 L, and we know that 1 litre = 1 dm³. So the volume of water is V = 310 dm³. Since this volume of water is the volume of water the cubical steel tank can contain, it equals the volume of the cubical steel tank.

We know that the volume of the cubical steel tank V = L³ where L is the length of side of the cube on the inside.

So the length of side of the cube L = ∛V = ∛310 dm³ = 6.77 dm = 6.77 dm × 1m/10 dm = 0.677 m ≅ 0.68 m

So, the smallest possible inside length of the tank is 0.68 m

(a) For the first 6.2 s, the car has velocity at time t given by

[tex]v(t)=13.5\dfrac{\rm m}{\rm s}+\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after 6.2 s, it attains a velocity of

[tex]v(6.2\,\mathrm s)=25.28\dfrac{\rm m}{\rm s}[/tex]

For any time t after 6.2 s, its velocity is given by

[tex]v(t)=25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

which tells us the velocity only falls from this point onward. This means the maximum speed is 25.28 m/s, or about 25.3 m/s.

(b) Solve for t (after 6.2 s) that makes v(t) = 0 :

[tex]25.28\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t=0[/tex]

[tex]\implies t\approx21.067\,\mathrm s[/tex]

It takes the car about 21.2 s to come to a rest, so the car travels a total of about 6.2 s + 21.2 s = 27.4 s.

(c) For the first 6.2 s, the car undergoes a displacement at time t of

[tex]x(t)=\left(13.5\dfrac{\rm m}{\rm s}\right)(6.2\,\mathrm s)+\dfrac12\left(1.9\dfrac{\rm m}{\mathrm s^2}\right)(6.2\,\mathrm s)^2[/tex]

[tex]\implies x\approx120.218\,\mathrm m[/tex]

For time t beyond 6.2 s, its displacement is

[tex]x(t)=120.218\,\mathrm m+\left(25.28\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

The car comes to a rest after 21.2 s (accelerating at a rate of -1.2 m/s^2), so that its total displacement is

[tex]x(21.2\,\mathrm s)\approx386.49\,\mathrm m[/tex]

so the car travels a total distance of about 387 m.

Answer:

An object opposes any change in its velocity, either to its direction or to its speed.

Explanation:

Edmentum Answer

Answer:

Yes

Explanation:

This is because the interaction between piece of paper and earth.is gravitational while that of piece of paper and rod is electrostatic

Yes, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

Therefore, the interaction between a piece of paper and the rod, stronger than the gravitational interaction between the piece of paper and the Earth.

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Answer:

3.38atm

Explanation:

Using data from the steam table we have that

Moles of water vapour = 907.19 / 18

= 50.4 moles

So

p1 = 30 psi = 30 x 0.68 = 2.04 atm

v1 = 4ft³= 113.2 L

Then from

PV= nRT

Then to find T we use

T1 = p1 V1 / n R

= 2.04 x 113.2 / 50.4 x 0.0821

= 55.8 K

Then to find volume two

v2 = 2v1 + v1

So

3 v1 = 339.6 K

The pressure two we use

P2 = n R T2 / V2

= 50.4 x 0.0821 x 277.6 / 339.6

So we have

= 3.38 atm =

Answer:

A. To find height of collision let's find

Speed of collision first

Using

Vf= √2g(h-g)

But for Harvey's ball we have

V²-2gy= 0

y=( v²/2g) so this is height of collision

B.

To find relative speed of the ball

Using V at y to find Vsr

Vs= √2g(h-v²/2g)

= √2gh- v²

The question is incomplete. Here is the complete question.

The map (in the attachment) shows Olivia's trip to the coffee shop. She gets on her bike at Loomis and then rides south 0.9mi to Broadway. She turns east onto Broadway, rides 0.8mi to where Broadway turns, and then continues another 1.4mi to the shop.

What is the magnitude of the total displacement of her trip?

Whta is the direction of the total displacement of her trip?

Answer: Magnitude = 2.6mi

              Direction: 54.65° east

Explanation: Displacement is the change in postition of a moving object.

There are a few ways to determine total displacement. For this case, the Perpendicular Components of a Vector method will be used.

For this method, total displacement is given by:

[tex]\Delta d_{t}=\sqrt{(\Delta d_{x})^{2}+(\Delta d_{y})^{2}}[/tex]

[tex]\theta=tan^{-1}(\frac{\Delta d_{y}}{\Delta d_{x}})[/tex]

[tex]\Delta d_{x}[/tex] is the x-component of total displacement and it is the sum of each individual x-components;

[tex]\Delta d_{y}[/tex] is the y-component of total displacement and it is the sum of each individual y-components;

θ is the angle the resulting displacement;

For Olivia's trip, there are no x-component of the first part and for the third part, the path she bikes is a hypotenuse of a right triangle. So, that right triangle's x-component is:

[tex]sin30=\frac{x}{1.4}[/tex]

[tex]\frac{1}{2} =\frac{x}{1.4}[/tex]

x = 0.7

Then,

[tex]\Delta d_{x}[/tex] = 0 + 0.8 + 0.7

[tex]\Delta d_{x}[/tex] = 1.5

Related to y, there are no y-component in the second part of Olivia's trip and for the third part:

[tex]cos30=\frac{y}{1.4}[/tex]

[tex]\frac{\sqrt{3} }{2} =\frac{y}{1.4}[/tex]

y = 1.21

Then,

[tex]\Delta d_{y}[/tex] = 0.9 + 0 + 1.21

[tex]\Delta d_{y}[/tex] = 2.11

Total displacement is

[tex]\Delta d_{t}=\sqrt{(1.5)^{2}+(2.11)^{2}}[/tex]

[tex]\Delta d_{t}=\sqrt{6.7021}[/tex]

[tex]\Delta d_{t}=[/tex] 2.6

Magnitude of Olivia's total displacement is 2.6mi

On the map, joining the initial and final points gives a vector pointing towards east at angle:

[tex]\theta=tan^{-1}(\frac{2.11}{1.5})[/tex]

[tex]\theta=tan^{-1}(1.41)[/tex]

θ = 54.65°

Direction of total displacement is 54.65° East.

Answer:

1.9 mi, 350.7°

Explanation:

If continued another 1.2 mi*

solving for x

0 + 0.8 + 1.2(cos30) = 1.83923048 --> 1.84

solving for y

-0.9 + 0 + 1.2(sin30) = -0.3

^negative because it is going downward

a) solving for magnitude

[tex]\sqrt{(1.84)^2+(-0.3)^2} = \sqrt{3.4756} = 1.86429611... = 1.9 mi[/tex]

b) solving for direction of total displacement

[tex]tan^-1 = (\frac{x}{y} ) \\tan^-1 = (\frac{-0.3}{1.84}) = -0.16 \\[/tex]

∘, measured counterclockwise from the eastward direction

360 - 0.16 = 359.84°

*replace any of the needed values in the equation, such as 1.2 mi to 1.4 mi

Answer:

The ball will be overtake superman after travelled distance 248.5 m.

Explanation:

Given that,

Speed of superman = 35 m/s

Height = 330 m

Mass of ball = 10 kg

Let the height at which they are at same position be S.

For superman,

We need to calculate the time

Using equation of motion

[tex]S=ut[/tex]...(I)

For ball,

[tex]S=u't+\dfrac{1}{2}gt^2[/tex]

From equation (I) and (II)

[tex]ut=u't+\dfrac{1}{2}gt^2[/tex]

[tex]ut=0+\dfrac{1}{2}\times g\times t^2[/tex]

[tex]u=\dfrac{1}{2}\times g\times t[/tex]

[tex]t=\dfrac{2u}{g}[/tex]

Put the value into the formula

[tex]t=\dfrac{2\times35}{9.8}[/tex]

[tex]t=7.1\ sec[/tex]

We need to calculate the distance travelled

Using formula of distance

[tex]d=v\times t[/tex]

Put the value into the formula

[tex]d=35\times 7.1[/tex]

[tex]d=248.5\ m[/tex]

Hence, The ball will be overtake superman after travelled distance 248.5 m.

Answer:

Rutherford

Explanation:

Because

Schrödinger I just took the unit review

Question:

What quantity do units represent in a value? A. Size B. Direction C. Magnitude D. Dimension

Answer:

D. Dimension

Explanation:

A unit is simply the measurement given to a dimension. In other words, units represent dimensions. For example, consider a ruler that is 20m long.

i. The value is 20m.

ii. The magnitude of the measurement is 20

iii. The unit of the measurement is meters(m) and that essentially represents the dimension (length) of the ruler.

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

Both chemical and physical changes involve energy transfer, so your friend is not correct.

A chemical change occurs when a new substance is formed or created through a process of chemical reaction which is reversible.

The addition or removal of heat energy can affect the rate of chemical reaction. This addition or removal is known as energy transfer process.

Examples of chemical changes include;

A physical change on the other hand doesn't involve formation of new substance and it is can be reversible or irreversible.

The addition or removal of heat energy can affect the rate of physical changes.

Examples of such physical changes include;

Thus, both process (chemical and physical changes) involve energy transfer, so your friend is not correct.

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Answer:

it gets colder the higher you go

Answer

To get critical pressure

We use

Pc = a/(27b²)

So

= 3.610/(27 X 0.0429²)

We have

= 72.7 atm

Critical temperaturewe

We use

Tc = 8a/27Rb

= 8 x 3.610/(27 x 0.0812 x 0.0429)

= 307 K

Critical volume

We use

Vc =3b =

3 x 0.0429

= 0.129L/mol

12,25 km/h

≈ 3,4 m/s

v = d/t

= 12250m/h

= 12,25km/h

or

v = d/t

= 12250m/h

1h = 60m×60s = 3600s

= 12250m/3600s

≈ 3,4 m/s

Answer:

The correct option is a

Explanation:

This question seeks to test a general rule in physics (on charges) which states that like charges repel but unlike charges attract. This means that, a negatively charged substance will repel or not attract another negatively charged material and the same applies to a positively charged substance also. However, a negatively charged substance will attract a positively charged material and vice versa, hence only a negatively charged rod will attract a positively charged hanging ball.

Answer:Newton’s law also states that larger bodies with heavier masses exert more gravitational pull, which is why those who walked on the much smaller moon experienced a sense of weightlessness, as it had a smaller gravitational pull. To help explain his theories of gravity and motion, Newton helped create a new, specialized form of mathematics.

Explanation:Here I dont know

Answer:

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction. Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction.

Explanation:

Answer:

Explanation:

Using the equation of motion S = ut + 1/2at² to get the height of drop where

u is the initial velocity of the object = 0m/s

a is the acceleration due to gravity = +9.81m/s² (downwards motion of object)

t is the time it takes the object to drop = 2secs

Substituting the given parameters into the formula to get the height of drop S, we will have;

S = 0(2)+1/2(9.81)(2)²

S = 0+9.81*2

S = 19.62m

Hence the object will drop at a distance of 19.62m

b.) To determine the time it will take a 5 kg object to drop the same distance, we will use the same formula S = ut+1/2at²

Since the 5kg object also drops at the same distance, then S = 19.62m

Substituting this values into the equation we will have;

19.62 = 0(t) + 1/2(9.81)t²

19.62 = 4.905t²

t² = 19.62/4.905

t² = 4

t =2secs

This shows that it will take 5kg object 2secs to fall from the  same distance. This means that no matter the mass of the object, it will take them the same time to fall at the same distance because they are all falling under the same influence of gravity.

Answer:

Calorimeter is used to measure heat in and represents that in units of joules per kelvin units J/˚C or kJ/K

A calorimeter is can be used to measure the amount of heat released or involved in a chemical reaction.

Whereas thermometer can only measures temperature or hotness of a substance. It cannot be used to measure the thermal rate or amount of heat energy of a reaction. Unit measurement used by thermometer is Celsius (°C).

Explanation:

Recall the formulas,

[tex]x_f=x_i+v_it+\dfrac12at^2[/tex]

[tex]v_f=v_i+at[/tex]

(a)

[tex]10.0\,\mathrm m=\left(19.0\dfrac{\rm m}{\rm s}\right)+\dfrac12\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

[tex]\implies \boxed{t\approx0.555\,\mathrm s}[/tex]

(b)

[tex]v_f=19.0\dfrac{\rm m}{\rm s}+\left(-3.50\dfrac{\rm m}{\mathrm s^2}\right)(0.555\,\mathrm s)[/tex]

[tex]\implies \boxed{v_f\approx17.1\dfrac{\rm m}{\rm s}}[/tex]

Answer: Heart rate provides an objective measurement of how hard your body is working. The higher the exercise intensity, the higher your heart rate will be.

Explanation:

Answer:

You're getting exercise and training your heart!

Explanation: