A horse gallops for 3.5 minutes around the raceing track. During this time it changes velocity from 543 min to 628 min. Calculate the acceleration of the horse during this 3.5min interval

Answer:

Mass, m = 71.42 kg

Explanation:

Given that,

Normal force acting on a box, F = 700 N

We need to find the mass of a box. Let it is m. Normal force acting on an object is balanced by its weight such that,

F = mg

Where m is the mass of the box

[tex]m=\dfrac{F}{g}\\\\m=\dfrac{700\ N}{9.8\ m/s^2}\\\\m=71.42\ kg[/tex]

So, the mass of the box is 71.42 kg.

Answer:

B. An opposing force caused by friction produced a lower acceleration than calculated

Explanation:

Answer:

An opposing force caused by friction produced a lower acceleration than calculated.

Explanation:

Answer:

I think thats the correct answer

Explanation:

Answer:

Explanation:

Given

the range = 5.00m (distance moved in horizontal direction)

Height = 0.150m

Required

Initial velocity of the dart

Using the formula for calculating range of a projectile;

R = U√2H/g

5 = U√2(0.15)/9.8

5 = U√0.0306

5 = 0.1749U

U = 5/0.1749

U = 28.59m/s

Hence the initial horizontal velocity of the dart is 28.59m/s

Answer:

28.59

Explanation:

khan academy

Answer:

Alpha particles are Helium nuclei consisting of 2 protons and two neutrons (mass number 4, atomic number 2, no electrons and so carry a 2+ charge).

Explanation:

Answer:

Height

Explanation:

the more height the more gravitational energy

Answer:

0.364

I believe... Good luck!

Answer:

Explanation:

The area of the base of the statue can be found by using the formula

[tex]a = \frac{f}{p} \\ [/tex]

f is the force

p is the pressure

From the question we have

[tex]a = \frac{1000}{20000} = \frac{1}{20} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

ΔT = 4.9°C

Explanation:

The thermal energy of the bar can be given as follows:

Thermal Energy = mCΔT

where,

m = mass of bar = 1 kg

C = specific heat capacity of aluminum = 1020 J/kg.°C

ΔT = Change in Temperature = ?

Therefore,

5000 J = (1 kg)(1020 J/kg.°C)ΔT

ΔT = (5000 J)/(1020 J/°C)

ΔT = 4.9°C

Answer:55,000 kg•m/s

Explanation:

Answer:

The arrow of the second archer standing on a high building will have more potential energy.

The arrow of the first archer standing on ground will have more kinetic energy.

Explanation:

POTENTIAL ENERGY:

The potential energy of an object depends upon its height, as given in the formula:

P.E = mgh

Hence, the arrow with the greater height will have more potential energy.

Therefore, the arrow of the second archer standing on a high building will have more potential energy.

KINETIC ENERGY:

The kinetic energy of an object depends upon its speed, as given in the formula:

K.E = (1/2)mv²

Hence, the arrow with the greater speed will have more kinetic energy energy. Since, the second arrow is stationary, it will have zero kinetic energy. But, the first arrow will have some K.E due to its speed.

Therefore, the arrow of the first archer standing on ground will have more kinetic energy.

(a)  The second archer standing at the top of the building will have more potential energy.

(b) The first archer on the ground level will have the more kinetic energy.

The potential energy of an object is the energy possessed by the object by virtue its position above above the ground.

P.E = mgh

where;

Thus, the second archer standing at the top of the building will have more potential energy.

(b) The kinetic energy of an object is the energy possessed by the object due to its motion.

K.E = ¹/₂mv²

where;

An object in motion, has zero velocity at maximum height but maximum velocity on the ground level.

Thus, the first archer on the ground level will have the more kinetic energy.

Learn more here:https://brainly.com/question/21866017

Answer:

C. It will remain constant because the total amount of thermal energy  plus chemical potential energy stays the same.

Explanation:

According to the law of conservation of energy, the total energy of this system will remain constant because the total amount of thermal energy plus chemical potential energy stays the same.

Answer:

it is hit by an external force

Explanation:

Answer:

ask questions that the findings bring up

Explanation:

hope this helps :)

Answer:

ask questions that the findings bring up

Answer:

Explanation:

The density of a substance can be found by using the formula

[tex]density = \frac{mass}{volume} \\ [/tex]

From the question we have

[tex]density = \frac{10.5}{19.3} \\ = 0.54404145...[/tex]

We have the final answer as

Hope this helps you

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

[tex]$m=\frac{V}{v}$[/tex]

[tex]$m=m_f+m_g$[/tex]

[tex]$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$[/tex]

[tex]$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$[/tex]       (taking the value of [tex]$v_g$[/tex] and [tex]$v_g$[/tex] at 200°C  )

[tex]$V=6.304 \times 10^{-3}$[/tex]

Now quality of vapor

[tex]$x=\frac{m_g}{m}$[/tex]

  [tex]$=3.377 \times 10^{-3}$[/tex]

Internal energy at state 1 can be found out by

[tex]$u_1=u_f+xu_{fg}$[/tex]

    [tex]$=850.65+3.377\times10^{-3}\times 1744.65$[/tex]

    = 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor [tex]$v_2=v_g \text { and }\ x=1$[/tex]

Tank is rigid, so volume of tank is constant.

[tex]$v_g=v_2=\frac{V}{m}$[/tex]

[tex]$v_g=\frac{6.304\times 10^{-3}}{1.4}$[/tex]

[tex]$v_g=4.502 \times 10^{-3} \ m^3 /kg$[/tex]

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

[tex]$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$[/tex]

   = 369.11° C

Internal energy at state 2

[tex]$u_2=2154.9 \ kJ/kg$[/tex]

Now power rating of the resistor

[tex]$P=\frac{m(u_2-u_1)}{t}$[/tex]

[tex]$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$[/tex]

  = 1.51 kW

Answer:

These deep-ocean currents are driven by differences in the water's density, which is controlled by temperature (thermo) and salinity (haline). This process is known as thermohaline circulation.

Explanation:

Answer:

A comparison of the dimensions of atom an ions that have the same number of electron but different nuclear charges is called isoelectronic series

Answer:

A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an isoelectronic series, shows a clear correlation between increasing nuclear charge and decreasing size

Answer:

option B is correct

Explanation:

neutralization is the reaction between acid and base to produce water and salt therefore option B is correct

Answer:

I think it's A and D.

Answer:

10m/s²

Explanation:

v=u+at

30=0+a(3)

∴a=10m/s²

Answer:

10m/s^2. Just curious, what kind of car is it? Is it a lamborghini

Explanation:

Answer:

no

Explanation:

because it will bind together in the molecular level

Answer:

1. potential 2. kinetic 3. potential 4. kinetic 5. kinetic 6. kinetic 7. kinetic 8. kinetic. potential

Answer:

Explanation:

Using the formula for calculating the maximum height in projectile

H = u²sin²theta/2g

u is the initial velocity = 18.3m/s

g is the acceleration due to gravity = 9.8m/s²

theta = 34°

Substitute

H = u²sin²theta/2g

H = 18.3²(sin34)²/2(9.8)

H = 334.89(0.5592)²/19.62

H = 5.337m

Hence the height of the cliff is 5.337m

Answer:

Drag, which is a force preventing the ball from going faster than it was thrown, this increases as velocity increases. Gravity, which Is pulling down on the ball.

Explanation:

Answer:

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Explanation:

We can estimate the absolute uncertainty by the definition of total differential. That is:

[tex]\Delta \rho \approx \frac{\partial \rho}{\partial m}\cdot \Delta m + \frac{\partial \rho}{\partial d}\cdot \Delta d + \frac{\partial \rho}{\partial l}\cdot \Delta l[/tex] (1)

Where:

[tex]\frac{\partial \rho}{\partial m}[/tex] - Partial derivative of the density with respect to mass, measured in [tex]\frac{1}{mm^{3}}[/tex].

[tex]\frac{\partial \rho}{\partial d}[/tex] - Partial derivative of the density with respect to diameter, measured in grams per cubic milimeter.

[tex]\frac{\partial \rho}{\partial l}[/tex] - Partial derivative of the density with respect to length, measured in grams per cubic milimeter.

[tex]\Delta m[/tex] - Mass uncertainty, measured in grams.

[tex]\Delta d[/tex] - Diameter uncertainty, measured in milimeters.

[tex]\Delta l[/tex] - Length uncertainty, measured in milimeters.

[tex]\Delta \rho[/tex] - Density uncertainty, measured in grams per cubic milimeters.

Partial derivatives are, respectively:

[tex]\frac{\partial \rho}{\partial m} = \frac{4}{\pi\cdot d^{2}\cdot l}[/tex] (2)

[tex]\frac{\partial \rho}{\partial d} = -\frac{8\cdot m}{\pi\cdot d^{3}\cdot l}[/tex] (3)

[tex]\frac{\partial \rho}{\partial l} = - \frac{4\cdot m}{\pi\cdot d^{2}\cdot l^{2}}[/tex] (4)

And we expand (1) as follows:

[tex]\Delta \rho \approx \frac{4\cdot \Delta m}{\pi\cdot d^{2}\cdot l} - \frac{8\cdot m\cdot \Delta d}{\pi\cdot d^{3}\cdot l}-\frac{4\cdot m\cdot \Delta l}{\pi\cdot d^{2}\cdot l^{2}}[/tex]

[tex]\Delta \rho \approx \left(\frac{4}{\pi\cdot d^{2}\cdot l}\right)\cdot \left(\Delta m -\frac{m\cdot \Delta d}{d}-\frac{m \cdot \Delta l}{l} \right)[/tex] (5)

If we know that [tex]d = 2\,mm[/tex], [tex]l = 25\,mm[/tex], [tex]m = 6.2\,g[/tex], [tex]\Delta m = \pm 0.1\,g[/tex], [tex]\Delta d = \pm 0.01\,mm[/tex] and [tex]\Delta l = \pm 0.1\,mm[/tex], then the absolute uncertainty is:

[tex]\Delta \rho \approx \pm\left[\frac{4}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)} \right]\cdot \left[(0.1\,g)-\frac{(6.2\,g)\cdot (0.01\,mm)}{2\,mm} -\frac{(6.2\,g)\cdot (0.1\,mm)}{25\,mm} \right][/tex]

[tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex]

And the expected density is:

[tex]\rho = \frac{4\cdot m}{\pi\cdot d^{2}\cdot l}[/tex] (6)

[tex]\rho = \frac{4\cdot (6.2\,g)}{\pi\cdot (2\,mm)^{2}\cdot (25\,mm)}[/tex]

[tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex]

The percentage uncertainty in his calculated value of density is:

[tex]\%e = \frac{\Delta \rho}{\rho}\times 100\,\%[/tex] (7)

If we know that [tex]\Delta \rho \approx \pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}}[/tex] and [tex]\rho \approx 78.941\times 10^{-3}\,\frac{g}{mm^{3}}[/tex], then the percentage uncertainty is:

[tex]\%e = \frac{\pm 5.628\times 10^{-4}\,\frac{g}{mm^{3}} }{78.941\times 10^{-3}\,\frac{g}{mm^{3}} }\times 100\,\%[/tex]

[tex]\%e = \pm 0.713\,\%[/tex]

The percentage uncertainty in his calculated value of density is [tex]\pm 0.713\,\%[/tex].

Answer:

Replication of the DNA must occur. Segregation of the "original" and its "replica" follow. Cytokinesis ends the cell division process. Whether the cell was eukaryotic or prokaryotic, these basic events must occur.