Which two fundamental forces are only attractive?
O A. Gravitational and strong nuclear
O B. Electromagnetic and weak nuclear
O C. Electromagnetic and strong nuclear
O D. Gravitational and weak nuclear

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:55,000 kg•m/s

Explanation:

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:

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:

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:

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:

I think thats the correct answer

Explanation:

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:

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:

I think it's A and D.

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:

The velocity the stone must be thrown with is 20 m/s.

Explanation:

Given;

time taken for the stone to reach it highest point, t = 2 s

acceleration due to gravity, g = 10 m/s²

The maximum height reached by the stone at the given time is calculated as;

[tex]h = \frac{1}{2}gt^2\\\\h = \frac{1}{2}*10*2^2\\\\h = 20 \ m[/tex]

Determine the initial of the stone;

v² = u² + 2gh

where;

v is the final velocity of the stone at maximum height = 0

u is the initial velocity of the stone = ?

0 = u² - 2(10 x 20)

0 = u² - 400

u² = 400

u = √400

u = 20 m/s

Therefore, the velocity the stone must be thrown with is 20 m/s.

Answer:

C

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:

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.

Answer:

1144.8

Explanation:

1 meter per second is equal to 3.6 kilometers per hour

Answer:

45 degrees

Explanation:

a projectile travels the farthest when it is launches at the angle of 45 degree.

Answer:

W = 1493.9 J = 1.49 KJ

Explanation:

The work done by the elevator on the object will be equal to the gain in is potential energy:

W = ΔP.E

W = mgΔh

where,

W = Work = ?

m = mass of object = 7.4 kg

g = 9.8 m/s²

Δh = gain in height = 20.6 m

Therefore,

W = (7.4 kg)(9.8 m/s²)(20.6 m)

W = 1493.9 J = 1.49 KJ

Answer:

Height

Explanation:

the more height the more gravitational energy

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:

no

Explanation:

because it will bind together in the molecular level

Answer:

0.364

I believe... Good luck!

Answer:

Speed is a description of how fast an object moves; velocity is how fast and in what direction it moves. In physics, velocity is speed in a given direction. When we say a car travels at 60 km/h, we are specifying its speed.

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:

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].