3. Thekinetic energy of an object of
mass, m moving with a velocity
of 5 ms-t is 25 J. What will be its
Toinetic energy when its velocity
is doubled? What will be its
Icinetic energy ruhen its velocity
is increased three times?​

Answers

Answer 1

Answer:

1. 100 J

2. 225 J

Explanation:

We'll begin by calculating the mass of the object. This can be obtained as follow:

Velocity (v) = 5 ms¯¹

Kinetic energy (KE) = 25 J

Mass (m) =?

KE = ½mv²

25 = ½ × m × 5²

25 = ½ × m × 25

25 = 25m / 2

Cross multiply

25m = 25 × 2

25m = 50

Divide both side by 25

m = 50 / 25

m = 2 Kg

1. Determination of the kinetic energy when the velocity is doubled.

Mass (m) = 2 Kg

Velocity (v) = double the initial velocity

= 2 × 5 ms¯¹

= 10 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 10²

KE = ½ × 2 × 100

KE = 100 J

2. Determination of the kinetic energy when the velocity increased three times.

Mass (m) = 2 Kg

Velocity (v) = three times the initial velocity

= 3 × 5 ms¯¹

= 15 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 15²

KE = ½ × 2 × 225

KE = 225 J


Related Questions

A cannon sitting on level ground is aimed at 45.0 degrees relative to the horizontal. It fires a test shot at a target located 100.0 meters away from the cannon on the same level ground. The test overshoots the target by 20.0 meters. Which of the following angles can the cannon be adjusted to to hit the target. You may neglect air resistance and assume the cannon always delivers the same initial velocity to the cannonball .

A. 35.9 deg
B. 49.1 deg
C. 28.2 deg
D. 52.8 deg
E. 22.7 deg

Answers

Answer:

C. 28.2 deg

Explanation:

The horizontal range of a projectile is given as:

[tex]R = \frac{v^2Sin2\theta}{g}[/tex]

where,

R = Range

v = speed

θ = angle of launch

g = acceleration due to gravity = 9.81 m/s²

First, we will find the launch speed (v) by using the initial conditions:

R = 120 m

θ = 45°

Therefore,

[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]

Now, consider the second scenario to hit the target:

R = 100 m

Therefore,

[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]

Hence, the correct option is:

C. 28.2 deg

Your mass is 80 kg on Earth. What would your MASS
be on Planet Roethler, which has twice as much
gravity as on Earth?
40 kg
160 kg
80 kg

Answers

Mass doesn't change. It doesn't matter where you are.

If your mass is 80 kg when you're on Earth, then your mass is 80 kg. It doesn't matter where you are, where you used to be, or where you're going tomorrow.

three condensers are connected in series across a 150 volt supply, the voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10^-8 c.calculate (a) the capacitance of each condenser (b)the effective capacitance of the combination

Answers

Answer:

(a) 1.5 nF, 1.2 nF, 1 nF

(b) 0.4 nF

Explanation:

V = 150 V

V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C

(a) C' = q/V' = 6 x 10^-8 / 40  = 1.5 x 10^-9  F

C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F

C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F

(b) The effective capacitance is

[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]  

How do space probes make it past the asteroid belt without crashing into asteroids?​

Answers

Answer:

The thing is space is really vast like really big so even though the asteroid belt looks really cramped it isn't. There's a lot of space between asteriods and using simple navigation and maneuvering, space probes can easily make it through without the threat of crashing.

Explanation:

what force to be required to accelerate a car of mass 120 kg from 5 m/s to 25m/s in 2s​

Answers

Answer:

[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]

Is it possible to get a body accelerated even if it is moving with uniform speed? justify.​

Answers

Answer:Yes, A body can have constant speed but still accelerate as in case of uniform circular motion. In uniform circular motion speed remains constant and direction of velocity changes with every point in the direction of tangent drawn from that point.

Explaination:

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Amy throws a softball through the air. What are the different forces acting on the ball while it’s in the air?
The softball experiences
force as a result of Amy’s throw. As the ball moves, it experiences
from the air it passes through. It also experiences a downward pull because of
.

Answers

Answer:

1.the friction of air, gravity2.gravity

Answer:

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.

Explanation:

Plato

Light strikes a smooth wooden tabletop.

What happens to the light after it is reflected?

The light rays bounce off the table and all move in the same direction.
The light rays bounce off the table and move in different directions.
The light rays pass through the table and all move in the same direction.
The light rays pass through the table and move in different directions.

Answers

Answer:

For smooth surface:The light rays bounce off the table and all move in the same direction.

The light will bounce off and continue in the direction it bounced.

What is displacement?
An objects overall change in position
The total measure of how far one travels
a change in direction

Answers

Answer:

An object overall change in position

Explanation:

ans maybe corre6

If the radius of curvature of a mirror is 15m and the distance of the object from the mirror is 10m. Find the distance of the image from the mirror and the magnification of the object in meter

Answers

Answer:

Data given.

focal length (f)=15m÷2=7.5m

Distance of the object(U)=10m

Image distance (v)=?

Magnification (M)=?

Solution:

From:

1/f=1/u+1/v

1/7.5=1/10+1/v=75

then v=75m

Magnification, M=u/v

=75/10=7.5

Then magnification=7.5

Answer:

v = 30 m and m = 3

Explanation:

Given that,

The radius of curvature of the mirror, R = 15 m

Focal length, f = 7.5 m

Object distance, u = -10 m

We need to find the image distance and the magnification of the object.

Using mirror's formula,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]

The magnification of the object in mirror is given by :

[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]

So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.

& What is meant by expansion effect of heating?​

Answers

Answer:

it's when heat demolished the object

Expansion. When heat is added to a solid, the particles gain energy and vibrate more vigorously about their fixed positions, forcing each other further apart. As a result expansion takes place. Similarly, the particles in a liquid or gas gain energy and are forced further apart.l

how much does a car weighr if each tyre cover of 4cm^2 and exert a pressure of 100000Nm^-2 on the ground it is physics questions ​

Answers

Answer:

Weight = 160 N

Explanation:

The weight of an object is a measure of the gravitational force on the object. So that weight is synonymous to the force applied by the object.

i.e Weight = mass x gravitational force

Also, the pressure of an object is the ratio of its force to area covered by the object. So that;

P = [tex]\frac{F}{A}[/tex]

Given that: area = 4 [tex]cm^{2}[/tex] (0.0004 [tex]m^{2}[/tex]), and pressure = 100 000 N[tex]m^{-2}[/tex].

Then;

F = PA

  = 100 000 N[tex]m^{-2}[/tex] x 0.0004 [tex]m^{2}[/tex]

F = 40 N

The force applied on the ground by each of the four tyres of the car is 40 N.

Thus, the weight of the car = 4 x 40 N

                                     = 160 N

What is matter made of.​

Answers

Answer:

Matter is made up of atoms

Answer:

Mater is made up of atoms.

Explanation:

Atoms come together to form molecules,which are the building blocks for all types of matter.

What process forms the Mid-Atlantic Ridge?
A. Radioactive decay
B. Seafloor spreading
C. Radiometric dating
D. Sediment formation

Answers

Answer:

B. seafloor spreading

Explanation:

divergent motion between the Eurasian and North American, and African and South American Plates. ... In this way, as the plates move further apart new ocean lithosphere is formed at the ridge and the ocean basin gets wider

geological society

Un alambre de plástico, aislante y recto mide 10 cm de longitud y tiene una densidad de carga de +150 nC/m, distribuidos de manera uniforme por toda su longitud. Se encuentra sobre una mesa horizontal. A) Encuentre la magnitud y la dirección del campo eléctrico que produce este alambre en un punto que está 8 cm directamente arriba de su punto medio. B) Si el alambre ahora se dobla para formar un círculo que se coloca aplanado sobre la mesa, calcule la magnitud y la dirección del campo eléctrico que produce en un punto que se encuentra 6 cm directamente arriba de su centro.

Answers

Answer:

English only

Explanation:

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here     cosθ   = h/ d   ⇒  cosθ = h/√h² + x²      dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻²  ;   dQ = λ×dx     λ = 150×10⁻⁹ C    h = 0.08 m

Then by substitution

Ey =  2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³          (2)

To solve the integral we make use of a change of variables

x = h × tanα     then   x² = h² ×tan²α   and  dx = h× sec²α dα

plugging that values in equation (2)

Ey  =  2×K×λ×h ∫  h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey  = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³            1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα             now we αneed to come back to our original variables:

as   x = h × tanα         tanα = x/h   then x is the opposite leg in a right triangle  and h the adjacent one then the hypothenuse is √ (h² + x²)         then    sin α = x/ √ (h² + x²)      

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey  = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 )   N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure  we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01      ( the length of the wire)  ⇒  r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h×  / ( √ r² + h²)³ ] × 10⁻²    N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C

5. A person standing on a 50 m cliff throws a rock upward at 26.5 m/s. How long will it take for the rock to reach the ground? Hint: determine the final velocity first.​

Answers

Answer:

53 seconds

Explanation:

26.5/50

0.53 seconds

It will take approximately 5.42 seconds for the rock to reach the ground from a 50 m cliff when thrown upward with an initial velocity of 26.5 m/s.

The time it takes for the rock to reach its highest point. We can use the equation:

Final velocity (v) = Initial velocity (u) + Acceleration (a) × Time (t)

0 = 26.5 - 9.8 × t

9.8 × t = 26.5

t = 26.5 ÷ 9.8

t = 2.71 seconds

Now, we double the time to account for the time it takes for the rock to reach its highest point and fall back down:

Total time = 2 × t

Total time ≈ 2 × 2.71

Total time = 5.42 seconds

Therefore, it will take approximately 5.42 seconds for the rock to reach the ground from a 50 m cliff when thrown upward with an initial velocity of 26.5 m/s.

To know more about the velocity:

https://brainly.com/question/34025828

#SPJ5

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... increased by a factor of 3, then the Fgrav is ______________ by a factor of _______. ... decreased by a factor of 4, then the Fgrav is ______________ by a factor of _______.

Answers

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=2M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=3M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]

so:

[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

Why are the largest craters we find on the Moon and Mercury so much larger than the largest craters we find on the Earth

Answers

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.

Answers

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

g is the symbol for grams.Exp (E) means exponential = 10um is the symbol for micrometer.

How much energy does it take to boil water for pasta? For a one-pound box of pasta
you would need four quarts of water, which requires 15.8 kJ of energy for every degree Celsius (°C) of temperature increase. Your thermometer measures the starting
temperature as 48°F. Water boils at 212°F.


a. [1 pts] How many degrees Fahrenheit (°F) must you raise the temperature?
b. [2 pts] How many degrees Celsius (°C) must you raise the temperature?
c. [2 pts] How much energy is required to heat the four quarts of water from
48°F to 212°F (boiling)?

Answers

Answer:

a. 164°F

b. [tex]91.\overline 1 \ ^{\circ} C[/tex]

c. [tex]140.\overline 4[/tex] kJ

Explanation:

The starting temperature of the water, T₁ = 48F

The temperature at which the water boils, T₂ = 212°F

a. The difference between the initial and the boiling water temperature, ΔT = T₂ - T₁

Therefore;

ΔT = 212°F - 48°F = 164°F

The temperature by which he temperature must be raised, ΔT = 164°F

b. 48°F = ((48 - 32)×5/9)°C = (80/9)°C = [tex]8.\overline 8 \ ^{\circ} C[/tex]

212°F = ((212 - 32)×5/9)°C = 100°C

∴ ΔT = 100°C - [tex]8.\overline 8 \ ^{\circ} C[/tex] = [tex]9.\overline 1 \ ^{\circ} C[/tex]

c. The heat capacity of the water = The heat required to increase four quartz of water by 1 °C = 15.8 kJ

∴ The heat required to raise four quartz of water by [tex]9.\overline 1 \ ^{\circ} C[/tex], ΔQ = 15.8 kJ/°C × [tex]9.\overline 1 \ ^{\circ} C[/tex] = [tex]140.\overline 4[/tex] kJ.

What is this?
Picture

Answers

Answer:

may be upside down alphabet :"T"

Explanation:

Two drums of the same size and same height are taken.
i)what will be the difference in liquid pressure on their bases if A of them is completely filled and B is half filled and kept at the same place.
ii) what will be the difference in liquid pressure on their bases if both A and B are filled with water completely but one of them is kept at nepal and another in india?why?
iii) what will be the difference in liquid pressure on their bases if A is filled with water and B is filled with salty water and kept at delhi in the same position?why?​

Answers

Answer:

i) The pressure acting on the base of B will be half the pressure acting on the base of A

ii) The pressure acting on the base of B will be the same as the pressure acting on the base of A

iii) The pressure on the base of drum A will be slightly less than the pressure on the base of drum B

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

[tex]h_{max}[/tex] = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If A is completely filled, we have [tex]h_A[/tex] = [tex]h_{max}[/tex]

Therefore, [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g

If B is half filled, we have, [tex]h_B[/tex] =  (1/2)·[tex]h_{max}[/tex]

[tex]P_B[/tex] = (1/2) × [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g

Therefore, [tex]P_B[/tex] = (1/2) × [tex]P_A[/tex]

The pressure acting on the base of B will be half the pressure acting on the base of A

ii) If both A and B are each filled with water (the same liquid), then the pressure on their bases will be [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g = [tex]P_B[/tex], the same, given that the acceleration due to gravity, g, is constant and the same in Nepal and India

iii) If A is filled with water, and B is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

[tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g <  [tex]P_B[/tex] =

The pressure on the base of drum A will be less than the pressure on the base of drum B.

a man is running on the straight road with the uniform velocity of3m/s.calculate its acceleration​

Answers

Answer:

9m is this the way calculator its acceleration

Statement:

A man is running on the straight road with the uniform velocity of 3 m/s.

To find out:

Acceleration produced by the man.

Solution:Given, the man is running with uniform velocity of 3 m/s.So the velocity did not change.It remains constant.We know, acceleration is the change of velocity per unit time.Since there is no change in velocity in this case, so there will be no acceleration.So, the acceleration produced by the man is zero.Answer:The acceleration produced by the man is zero.

A gas is enclosed in a confainer fitted with a piston of cross sectional area 0.10 the pressureof the gas is maintained in 8000 when hat is slowlh transferred the piston is pushed up through a distance of 4.0cm If 42j of heat is transferred to the system during expansion wht is the change im internal energy of th system

Answers

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

The power of the kettle was 1.5 kW. The 0.2kg heating element took 5 seconds to heat from 20 °C to 100 °C. Calculate the specific heat capacity of water using this information.

Answers

Answer:

Specific heat capacity, c = 468.75 J/Kg°C

Explanation:

Given the following data;

Power = 1.5 kW to Watts = 1.5 * 1000 = 1500 Watts

Time = 5 seconds

Mass = 0.2 kg

Initial temperature = 20°C

Final temperature = 100°C

To find specific heat capacity;

First of all, we would have to determine the energy consumption of the kettle;

Energy = power * time

Energy = 1500 * 5

Energy = 7500 Joules

Next, we would calculate the specific heat capacity of water.

Heat capacity is given by the formula;

[tex] Q = mcdt[/tex]

Where;

Q represents the heat capacity or quantity of heat. m represents the mass of an object. c represents the specific heat capacity of water. dt represents the change in temperature.

dt = T2 - T1

dt = 100 - 20

dt = 80°C

Making c the subject of formula, we have;

[tex] c = \frac {Q}{mdt} [/tex]

Substituting into the equation, we have;

[tex] c = \frac {7500}{0.2*80} [/tex]

[tex] c = \frac {7500}{16} [/tex]

Specific heat capacity, c = 468.75 J/Kg°C

3.00 m^3 of water is at 20.0°C.
If you raise its temperature to
60.0°C, by how much will its
volume expand?
Water
B = 207•10-6 0-1
(Unit = m^3)

Answers

Answer:

[tex]\triangle V = 0.02484m^3[/tex]

Explanation:

Given

[tex]V_1 = 3.00m^3[/tex] --- initial volume

[tex]T_1 = 20.0^oC[/tex] --- initial temperature

[tex]T_2 = 60.0^oC[/tex] --- final temperature

[tex]\gamma = 207*10^{-6[/tex] ---  coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

[tex]\triangle V = \gamma * V_2 * (T_2 - T_1)[/tex]

So, we have:

[tex]\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)[/tex]

[tex]\triangle V = 207 * 10^{-6} * 3.00 * 40.0[/tex]

[tex]\triangle V = 0.02484m^3[/tex]

The volume will expand by [tex]0.02484m^3[/tex]

Sí un auto viaja a 8m/s determine Tiempo en llegar a 200km de distancia Distancia que recorre en 40 minutos

Answers

Answer:

a. Tiempo = 25000 segundos

b. Distancia = 19200 metros

Explanation:

Dados los siguientes datos;

Velocidad = 8 m/s

Distancia = 200 km a metros = 200 * 1000 = 200,000

Para encontrar el tiempo para cubrir la distancia anterior;

Tiempo = distancia/velocidad

Tiempo = 200000/8

Tiempo = 25000 segundos

b. Para encontrar la distancia recorrida en 40 minutos;

Tiempo = 40 minutos a segundos = 40 * 60 = 2400 segundos

Distancia = velocidad * tiempo

Distancia = 8 * 2400

Distancia = 19200 metros

Please help me with this!!!​

Answers

Answer:

Its all about figuring out what numbers you times by

Explanation:

soooooooooo just x the numbers until you get it right, and i'm guessing your in a school soooo ask your teacher aswell :)

An ice skater pushes harder with her legs and begins to move faster. Which two laws best describes this

Answers

Answer:

Newton' second law  and third law describes the situation.

Explanation:

According to the Newton's second law, the force applied on a body is proportional to the rate of change of momentum of the body.

According to the Newton's third law, for every action there is an equal and opposite reaction.

When ice skater pushes harder means more force is applied so he moves fast and more be the action force more be the reaction force.

Thus, Newton' second law  and third law describes the situation.

Which would be used to measure the distance between the earth and a planet ,meter ruler or a measuring tape? Why?

Answers

Answer:

parallax

Due to foreshortening, nearby objects show a larger parallax than farther objects when observed from different positions, so parallax can be used to determine distances. To measure large distances, such as the distance of a planet or a star from Earth, astronomers use the principle of parallax.

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