A 64.1 kg runner has a speed of 3.10 m/s at one instant during a long-distance event.(a) What is the runner's kinetic energy at this instant?
KEi = _________________J
(b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?
KEf/KEi=______________

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:

So that vehicles do not topple or skid off the road.

Explanation:

As a vehicle is negotiating a curve, two equal and opposite forces act on the vehicle; centripetal force which keeps it moving and centrifugal force which tends to throw it out of the path.

In order to avoid skidding off the road curved roads are banked. Banking a curved surface provides the centripetal force that points towards the center of the road hence the vehicle or car does not skid off or topple.

Answer:

The maximum speed of turn on the given circular track is 16.27 m/s.

Explanation:

Given;

mass of the car, m = 2200 kg

radius of the track, r= 30 m

coefficient of static friction between the tires and the road, μ = 0.9

The net vertical force on the car = N = mg

The net horizontal force on the car = Centripetal force

The coefficient of static friction is given as;

[tex]\mu = \frac{F_c}{N} \\\\[/tex]

[tex]F_c = \mu N\\\\\frac{mv^2}{r} = \mu mg\\\\ \frac{v^2}{r}= \mu g\\\\v^2 = \mu gr\\\\v = \sqrt{\mu gr}[/tex]

where;

v is the maximum speed of turn

[tex]v = \sqrt{\mu gr} \\\\v = \sqrt{0.9 \times 9.8 \times 30 } \\\\v = 16.27 \ m/s[/tex]

Therefore, the maximum speed of turn on the given circular track is 16.27 m/s.

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.

Explanation:

p.e =mgh

given: m=2350g=2.35kg h=20 g=9.8m/s

p.e=mgh

=2.35kg×20.0m×9.8

=460.6j

I am not sure

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

Answer:

the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance. 

The correct answer is 195.6 N

Explanation:

Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:

F = 120 kg x 1.63 m/s2

F= 195.6 N

Answer:

mass =22.4kg

force=83.1N

a=?

f=ma

a=f/m

a=83.1/22.4

a=3.70m/s^2

Answer:

50000 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of bullet = 0.050 kg

velocity (v) = 400 m/s

Distance (s) = 0.080 m

Force (F) =?

Next, we shall determine the acceleration of the bullet. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 400 m/s

Distance (s) = 0.080 m

Acceleration (a) =?

v² = u² + 2as

400² = 0 + (2 × a × 0.08)

160000 = 0 + 0.16a

160000 = 0.16a

Divide both side by 0.16

a = 160000 / 0.16

a = 1×10⁶ m/s²

Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:

Mass (m) of bullet = 0.050 kg

Acceleration (a) of bullet = 1×10⁶ m/s²

Force (F) =?

F = ma

F = 0.050 × 1×10⁶

F = 50000 N

Thus, the bullet exerted a force of 50000 N on the target.

Answer:

10m

Explanation:

let's take the acceleration as a constant throughout the complete motion...

therefore first let's find the acceleration

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ 2 = \frac{1}{2} a {1}^{2} \\ a = 4m {s}^{ - 2} [/tex]

then we have to find v1

apply V = u + at

v = 4×1

= 4ms^-1

lets find the distance travel by the object DURING THE TIME INTERVAL 1-2

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 4 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s= 6m [/tex]

then let's find the V2

[tex] {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = {4 }^{2} + 2 \times 4 \times 6 \\ {v}^{2} = 64 \\ v = \sqrt{64} = 8m {s}^{ - 1} [/tex]

[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 8 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s = 10m [/tex]

The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.

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, an object released from rest at time t = 0 slides down a frictionless incline a distance of 2 m during the time interval from t=0 seconds to t = 1 seconds.

2 = ut + 1/2*a*t²

2 = 0 + 0.5×a×1²

a = 2 / 0.5

a = 4 meters / second²

Now to find the velocity after the 2 seconds,

v = u + at

v = 0 + 4×2

v = 8 m/s

Now by using the second equation of motion,

S = ut + 1/2×a×t²

S = 8×1 + 0.5×4×1²

S = 8 + 2

S = 10 meters

Thus, The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.

To learn more about equations of motion here,

brainly.com/question/5955789

#SPJ5

Explanation:

Using Kinematics,

we have a = (v - u) / t.

Therefore a = (36m/s - 22m/s) / 5s = 2.8m/s².

Answer:Nuclear potential energy is the potential energy of the particles inside an atomic nucleus

Answer:

Peninsula

Explanation:

Just did the Ed puzzle and got it right

Answer:

0.426 volts

Explanation:

It is given that,

The radius of a circular loop, r = 11.2 cm = 0.112 m

An elastic conducting material is stretched into a circular loop.

It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s

We need to find the emf induced in the loop at that instant.

[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{d}{dt}(BA)\\\\=\dfrac{d}{dt}(\pi r^2 B)\\\\=\pi B\dfrac{d}{dt}(r^2)\\\\=2\pi B r\dfrac{dr}{dt}\\\\=2\pi \times 0.88\times 0.112\times 0.688\\\\=0.426\ V[/tex]

So, the magnitude of induced emf is 0.426 volts.

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

Answer:

English please?

Explanation:

Explanation:

K.E=1/2mv^2

m=1.5×10^5

v=70

K.E=1/2×1.5×70

K.E=52.5×10^5

Answer:

3 cm³

Explanation:

Density of the cork = 0.6 g/cm³

Density of water = 1 g/cm³

Volume of cork = 5 cm³

We all know that the formula for density is given as

Density = mass/volume,

The mass of the cork is

Mass = density * volume

Mass = 0.6 * 5

Mass = 3 gram

Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is

5 * 0.6 = 3 cm³

The volume of cork is below the water 3 cm³.

Given data:

The density of Cork is,  [tex]\rho = 0.60 \;\rm cm^{3}[/tex].

The density of water is, [tex]\rho_{w} = 1.0 \;\rm g/cm^{3}[/tex].

The total volume of piece of cork is,  [tex]V = 5.0 \;\rm cm^{3}[/tex].

In the given problem, we can use the concept of density. The mass of substance occupied per unit volume is known as Density of substance. The expression for the density is given as,

Density = mass/volume,

And, the mass of the cork is

Mass = density * volume

Mass = 0.6 * 5

Mass = 3 gram

Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is,

= 5 * 0.6 = 3 cm³

Thus, we can conclude that the volume of cork is below the water 3 cm³.

learn more about the concept of density here:

https://brainly.com/question/21667661

Answer:

dosimetry

Explanation:

dosimetry is the determination of energy imparted to matter by radiation

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:

It will not hit.

Explanation:

In this problem, we need to tell that was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s.

When it will hit the target, final velocity, v = 0

Using third equation of motion as follows :

[tex]v^2-u^2=2as[/tex]

Here, a =-g

[tex]s=\dfrac{u^2}{2g}\\\\s=\dfrac{(20)^2}{2\times 9.8}\\\\=20.40\ m[/tex]

The target is in a straight line 50m away on the x-axis. As 50m is far from 20.40 m and that’s why it won’t hit.

Answer:

1:36,056,338

Explanation:

First thing we do is convert the diameter of the earth from kilometers into centimeters. Thus, we have

12800 km = 12800 * 100000 cm

12800 km = 1280000000 cm

Then we have diameter of the globe to be 35.5 cm.

To get the scale, what we do is divide the diameter of the earth by that of the globe, and as such we have

1280000000 / 35.5 =

36056338.

Therefore, the scale of the globe is 1:36,056,338

For every 1 cm on the globe, it is 36,056,338 cm or 360.6 km on earth

The scale of this globe is equal to 1 : 36,056,338.

Given the following data:

In Science, a scale can be defined as a ratio of the distance on a map to the actual (corresponding) distance on planet Earth.

A globe refers to a scale model of planet Earth that accurately depicts various geographic information such as distance, circumference, area, etc.

To determine the scale of this globe:

First of all, we would convert the value of Earth diameter in kilometers to centimeters as follow:

Conversion:

1 kilometer = 100,000 centimeter

12,800 kilometer = [tex]12800 \times 10^5 = 128 \times 10^7\;centimeters[/tex]

Now, we can calculate the scale of the globe by using this formula:

[tex]Distance = \frac{Earth\;diameter}{Globe\;diameter}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Distance = \frac{128 \times 10^7}{35.5}[/tex]

Distance = 36,056,338.03 ≈ 36,056,338 centimeters.

Scale = 1 : 36,056,338.

Read more on globe here: https://brainly.com/question/5659485

Answer:

small marble  

Explanation:

A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.

In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

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: It is the color of the container

Explanation:

The red at the bottom is the color of the bottom of the container. It is not part of the experiment and is not liquid.

The red liquid that is part of the experiment does indeed have the lowest density of the liquids which is why it is floating at the top of al the liquids with the only thing floating on top of it being the blue cube which has a lower density than it.

Answer: 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:

inversely proportional to

Answer:

6 Fahrenheit, converted over to Celsius, would be -14.444444

Answer:

A. True

Explanation:

Type la supernova can be described as a phenomenon whereby two stars are in orbit with one another .

Recently , two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances that the universe is accelerating while it expands.