Two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances, this being that the universe is accelerating while it expands.
A. True
B. False

Answer:

(c) moves toward the shore.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = mass * velocity [/tex]

The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.

In this scenario, a boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, away from shore. As a consequence, the raft moves toward the shore because the momentum of the closed system (boy and raft) has a zero magnitude and would remain constant.

Answer:

Surface tension is the downward force acting on the surface of liquid due to presence of inter molecular forces or cohesive forces between the particles of liquid.

Surface tension decreases with increase in temperature as the forces among particles decrease due to increase in kinetic energy and thus the cohesive nature decreases and thus surface tension also decreases.

Surface tension may decrease or increase with increase in soluble impurities .Insoluble impurities decrease the surface tension.

Answer:

The road bank angle is 16.38⁰.

Explanation:

radius of curvature of the road, r = 50 m

allowable speed of car on the road, v = 12 m/s

The bank angle is calculated as;

[tex]\theta = tan^{-1} (\frac{v^2}{gr} )[/tex]

where;

θ is the road bank angle

g is acceleration due to gravity = 9.8 m/s²

[tex]\theta = tan^{-1} (\frac{v^2}{gr} )\\\\\theta = tan^{-1} (\frac{12^2}{9.8 \times 50} )\\\\\theta = tan^{-1} ( 0.2939)\\\\\theta = 16.38 ^0[/tex]

Therefore, the road bank angle is 16.38⁰.

Answer:

  15√2 N

Explanation:

The acceleration is given by ...

  a = F/m = 5t/5 = t . . . . meters/second^2

The velocity is the integral of acceleration:

  v = ∫a·dt = (1/2)t^2

This will be 9 m/s when ...

  9 = (1/2)t^2

  t = √18 . . . . seconds

And the force at that time is ...

  F = 5(√18) = 15√2 . . . . newtons

Answer:

The value is [tex]R = 0.321 \ \Omega[/tex]

Explanation:

From the question we are told that

   The diameter is  [tex]d = 8.252 \ mm = 0.008252 \ m[/tex]

    The length of the wire is  [tex]l = 1.0 \ km = 1000 \ m[/tex]

   Generally the cross sectional area of the copper wire is mathematically represented as

           [tex]A = \pi * \frac{d^2}{4}[/tex]

=>        [tex]A = 3.142 * \frac{ 0.008252^2}{4}[/tex]

=>         [tex]A = 5.349 *10^{ - 5} \ m^2[/tex]

Generally the resistance is mathematically represented as

      [tex]R = \frac{\rho * l }{A }[/tex]

Here [tex]\rho[/tex] is the resistivity of copper with the value  [tex]\rho = 1.72*10^{-8} \ \Omega \cdot m[/tex]

=>    [tex]R = \frac{1.72 *10^{-8} * 1000 }{5.349 *10^{ - 5} }[/tex]

=>    [tex]R = 0.321 \ \Omega[/tex]

Answer:

a. [tex]m' = \frac{m}{4}[/tex]

b. [tex]m' = \frac{m}{9}[/tex]

Explanation:

The frequency of a harmonic oscillator is given by the following formula:

[tex]\omega = \sqrt{\frac{k}{m}}[/tex]   ----------------- equation (1)

a.

In order to double the frequency of this oscillator:

ω' = 2ω

m' = ?

Therefore,

[tex]\omega ' = 2\omega = \sqrt{\frac{k}{m'}}[/tex]

using equation (1):

[tex]2 \sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{4}{m} = \frac{1}{m'}[/tex]  

[tex]m' = \frac{m}{4}[/tex]

a.

In order to triple the frequency of this oscillator:

ω' = 3ω

m' = ?

Therefore,

[tex]\omega ' = 3\omega = \sqrt{\frac{k}{m'}}[/tex]

using equation (1):

[tex]3\sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{9}{m} = \frac{1}{m'}[/tex]  

[tex]m' = \frac{m}{9}[/tex]

A) To double the Frequency of a harmonic oscillator ;

Divide the mass by four       i.e. m₁ = m / 4

B) To triple the frequency of a harmonic oscillator :

Divide the mass by nine (9)     i.e.  m₂ = m / 9

Given that The frequency of a harmonic oscillator is expressed as  

[tex]w = \sqrt{\frac{k}{m} }[/tex]  -- ( 1 )

A) Doubling the frequency

[tex]2w = \sqrt{\frac{k}{m_{1} } }[/tex]  ---- ( 2 )

Applying equation ( 1 ) and ( 2 )

[tex]2\sqrt{\frac{k}{m} } = \sqrt{\frac{k}{m_{1} } }[/tex]

squaring both sides

( 4 / m )  =  1 / m₁

∴ m₁ ( new mass ) = m / 4

B) Tripling the frequency

3w = [tex]\sqrt{\frac{k}{m_{2} } }[/tex]    ---- ( 3 )

applying equation ( 1 ) and ( 3 )

[tex]3 \sqrt{\frac{k}{m} } = \sqrt{\frac{k}{m_{2} } }[/tex]  

squaring both sides

( 9 / m ) =  1 / m₂

∴ m₂ = m / 9

Hence we can conclude that To double the Frequency of a harmonic oscillator   m₁ = m / 4  and To triple the frequency of a harmonic oscillator : m₂ = m / 9

Learn more : https://brainly.com/question/20050933

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

Answer:

Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.

Probably D:  Soccer ball was affected by air resistance more than the golf ball.

Explanation:

Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.

In a vacuum, (no friction) both objects fall at the same rate regardless of mass.

Answer:

T = 365.58 K

Explanation:

Given that,

The concentration of solution, C = 0.750M

Osmotic pressure, P = 22.5 atm

We need to find the temperature of the solution.

The formula for the osmotic pressure is given by :

[tex]P=CRT[/tex]

Where

R is gas constant, [tex]R=0.08206\ L\ atm/mol-K[/tex]

[tex]T=\dfrac{P}{CR}\\\\=\dfrac{22.5}{0.75\times 0.08206}\\\\=365.58\ K[/tex]

So, the temperature of the solution is 365.58 K.

Explanation:

because magnet attract opposite sides. like north and south.

Answer:

Its because the magnet has both north and south pole, so when suspended it turns its south pole towards southern hemisphere and the north pole towards northern hemisphere

Hope it helped u,

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Answer:

A volcano fed by highly viscous magma is likely to be a greater threat to life and property than a volcano supplied with very fluid magma because with high viscous magma gas is trapped more in the magma so the gas will build up and then eventually explode, whereas with fluid magma the gas can escape allowing the magma.

HOPE THIS HELPS!!!

Explanation:

Less fluid magma done great damaged to the property and life as compared to highly viscus magma.

No, volcanoes that are fed by highly viscous magma are not a greater threat to life and property than volcanoes supplied with very fluid magma because the highly viscous magma can't move to a large distance due to its large viscosity.

While on the other hand, those volcanoes that supplied with very fluid magma do great damaged to the property due to its easily flowing on the surface of earth so we can conclude that less fluid magma done great damaged to the property and life as compared to highly viscus magma.

Learn more about magma here: https://brainly.com/question/23661578

Answer:

C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.

Explanation:

Hooke's law states that the force needed to extend or compress a spring by a distance is proportional to that distance. If is given as:

F = ke, where F is the force applied, k is spring constant and e is the extension.

If a force f is applied to a spring with a spring constant k and by a distance stretched (x) then:

k = F / x

For half the spring, if the same force F is applied, the stretch would be half (x/2), hence the spring constant C is:

C = F / (x/2)

C = 2 (F / x) = 2 * spring constant of original spring

Answer:

a) The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Explanation:

a) By Newton's laws we construct the following equations of equilibrium. Please notice that both the glider and the jet experiments has the same acceleration:

Jet

[tex]\Sigma F = F - T = m_{J}\cdot a[/tex] (1)

Glider

[tex]\Sigma F = T = m_{G}\cdot a[/tex] (2)

Where:

[tex]F[/tex] - Thrust of jet engines, measured in newtons.

[tex]T[/tex] - Tension in the cable connecting the jet and glider, measured in newtons.

[tex]m_{G}[/tex], [tex]m_{J}[/tex] - Masses of the glider and the jet, measured in kilograms.

[tex]a[/tex] - Acceleration of the glider-jet system, measured in meters per square second.

If we know that [tex]m_{G} = 260\,kg[/tex], [tex]m_{J} = 1,940\,kg[/tex] and [tex]a = 2.20\,\frac{m}{s^{2}}[/tex], then the solution of this system of equations:

By (2):

[tex]T = (260\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)[/tex]

[tex]T = 572\,N[/tex]

By (1):

[tex]F = T+m_{J}\cdot a[/tex]

[tex]F = 572\,N+(1,940\,kg)\cdot \left(2.20\,\frac{m}{s^{2}} \right)[/tex]

[tex]F = 4840\,N[/tex]

The magnitude of the thrust provided by the jet's engines is 4840 newtons.

b) The magnitude of the tension in the cable connecting the jet and glider is 572 newtons.

Answer: the child's centripetal acceleration=24.50 m/s²

Explanation:

Given that mass of child= 50 kg

radius of merry go round= 2.25m

angular speed = 3.30 rad/s

Centripetal Acceleration  = v²/ r

  But  V= ωr

So Centripetal Acceleration  = v²/ r =  (ωr)²/ r

=(3.30)² x  (2.25)²/ 2.25 = (3.30)² x  2.25

=24.5025m/s²

=24.50 m/s²

Answer:

If a population exceeds carrying capacity, the ecosystem may become unsuitable for the species to survive. If the population exceeds the carrying capacity for a long period of time, resources may be completely depleted.

Answer:

Efriction = 768.23 [kJ]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Where it tells us that the energy of a system plus the work applied or performed by that system, will be equal to the energy in the final state. We have two states the initial at the time of the balloon jump and the final state when the parachutist lands.

We must identify the types of energy in each state, in the initial state there is only potential energy, since the reference level is in the ground, at the reference point the potential energy is zero. At the time of landing the parachutist will only have potential energy, since it reaches the reference level.

The friction force acts in the opposite direction to the movement, therefore it will have a negative sign.

[tex]E_{pot}-E_{friction}=E_{kin}[/tex]

where:

[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]

m = mass = 56 [kg]

h = elevation = 1400 [m]

v = velocity = 5.6 [m/s]

[tex](56*9.81*1400)-E_{friction}=\frac{1}{2}*56*(5.6)^{2}\\769104 -E_{friction}= 878.08 \\E_{friction}=769104-878.08\\E_{friction}=768226[J] = 768.23 [kJ][/tex]

Answer:

The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.

Explanation:

A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  [tex]D = 0.59 \ m[/tex]    

Explanation:

From the question we are told that

      The best resolution is  [tex]\theta = 0.3 \ arcsecond[/tex]

       The  wavelength is  [tex]\lambda = 700 \ nm = 700 *10^{-9 } \ m[/tex]

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       [tex]x = \frac{0.3}{60 }[/tex]

=>    [tex]x = 0.005 \ arcminutes[/tex]

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       [tex]z = \frac{0.005}{60 }[/tex]

=>      [tex]z = 8.333 *10^{-5} \ degree[/tex]

Converting to radian  

           [tex]\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian[/tex]

Generally the resolution is mathematically represented as

            [tex]\theta = \frac{1.22 * \lambda }{ D}[/tex]

=>    [tex]D = \frac{1.22 * \lambda }{\theta }[/tex]

=>     [tex]D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }[/tex]    

=>     [tex]D = 0.59 \ m[/tex]    

Answer:

a

  [tex]F = 0.0566 \ N[/tex]  

b

   [tex]t = 6.147 \ s[/tex]

Explanation:

From the question we are told that

     The distance travel in  4.22 s is  [tex]s = 1.44 \ m[/tex]

     The mass of the cart plus the fan is  [tex]m = 350 \ g = 0.35 \ kg[/tex]

Generally from kinematic equation we have that

        [tex]s = ut + \frac{1}{2} * a * t^2[/tex]

Here  u is the initial  velocity with value  [tex]u = 0 \ m/s[/tex]

So  

         [tex]1.44= 0 * t + \frac{1}{2} * a * 4.22^2[/tex]      

=>      [tex]a = 0.1617 \ m/s^2[/tex]

Generally the net force is  

         [tex]F = m * a[/tex]

=>      [tex]F = 0.35 * 0.1617[/tex]  

=>      [tex]F = 0.0566 \ N[/tex]  

Gnerally the new mass of the cart plus the fan is  [tex]M = 656 \ g = 0.656 \ kg[/tex]

    The distance considered is [tex]s_1 = 1.63 \ m[/tex]

     Generally the new acceleration of the cart is mathematically represented as

        [tex]F = M * a_1[/tex]

=>      [tex]a_1 = \frac{F}{M}[/tex]

=>      [tex]a_1 = \frac{0.0566}{0.656}[/tex]

=>      [tex]a_1 = 0.08628 \ m/s^2[/tex]

Gnerally from kinematic equation we have

          [tex]s = ut + \frac{1}{2} * a_1 * t ^2[/tex]

Here u  is the initial velocity and the value is zero because it started from rest  

=>       [tex]1.63 = 0 * t + \frac{1}{2} * 0.08628* t ^2[/tex]

=>        [tex]t = 6.147 \ s[/tex]

Answer:

The leaves have a charge in each experiment, but the sign of the charge cannot be determined.

Explanation:

In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.

Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.

Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.

Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.

Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.

Answer:

(i) Relative velocity of B w.r.t A= Sum of speeds of trains

=54+90

=144kmph

(ii)Relative velocity of B w.r.t Ground(G)=v

B/G

=−90kmph

v

G

=0

Relative velocity of ground(G) w.r.t B =v

G/B

=v

G

−v

B/G

v

G/B

=0−(−90)

v

G/B

=90kmph

Answer:

The answer to your question is given below

Explanation:

To solve this problem, we'll assume that the plane is initially at rest.

Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0

Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:

Force (F) = 5000 N

Distance (s) = 500 m

Energy (E) =?

E = F × s

E = 5000 × 500

E = 2500000 J

Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J

Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:

Initial kinetic energy (KE₁) = 0

Final kinetic energy (KE₂) = 2500000 J

Change in kinetic energy (ΔKE) =?

ΔKE = KE₂ – KE₁

ΔKE = 2500000 – 0

ΔKE = 2500000 J

Hence, the change in the kinetic energy of the plane is 2500000 J.

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 15.5 × 24.2

We have the final answer as

Hope this helps you

Answer:

Answer:

a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.

Explanation:

Answer:

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Answer:

That is not meant to be red, it is the bottom of the beaker

That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.

Answer:

a) 4.9m/s²

b) 2.18m/s²

Explanation:

a) According to Newton's second law of motion

\sum Fx = ma

Fm-Ff = ma

Fm is the moving force = Wsin theta

Ff is the frictional force = 0N (frictionless plane)

m is the mass

a is the acceleration

Substituting into the formula

Fm -Ff = ma.

Wsintheta = ma

mgsintheta = ma

gsintheta = a

a = 9.8sin30°

a = 9.8(0.5)

a = 4.9m/s²

Hence the acceleration of the block if the plane is frictionless is 4.9m/s²

b) Let the coefficient if friction given be 0.32

From the formula

Fm-Ff = ma

mgsintheta - nmgcostheta = ma

gsintheta - ngcostheta = a

9.8(sin30)-0.32(9.8)cos30 = a

4.9-2.72 = a

a = 2.18m/s²

Hence the acceleration if the coefficient of kinetic friction is 0.32 is 2.18m/s²