The air around a pool and the water in the pool receive equal amounts of energy from the sun. Why does the air experience a greater increase in temperature than does the water?

A.
The specific heat of water is less than that of air.

B.
The specific heat of water is greater than that of air.

C.
The latent heat of water is less than that of air.

D.
The latent heat of water is greater than that of air.

Answer: Dynamic - Static Flexibility

Explanation:

Answer:

go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school

Answer:

70 m/s.

Explanation:

Given that,

Initial speed, u = 40 m/s

Acceleration = 2.5 m/s²

Time, t = 12 s

We need to find criminal's new  speed. Let it is v. Using equation of motion to find it as follows :

v = u +at

Substitute all the values

v = 40 + 2.5(12)

v = 70 m/s

So, the new speed is 70 m/s.

Answer:

5 sig figures

Explanation:

Answer:

heat

Explanation:

Answer:

4. atmosphere and geosphere

5. atmosphere and hydrosphere

6. hydrosphere and geosphere (? not sure about this one sorry)

7. hydrosphere and geosphere

8. biosphere and geosphere

Odpowiedź:

0,5 m / s²

Wyjaśnienie:

Jeśli się uwzględni:

Masa samochodu (m) = 2000 kg

Wartość siły oporu podczas holowania = 400N

Lina o maksymalnej sile może wytrzymać = 1400N

Maksymalne przyspieszenie = siła / masa samochodu

Siła = maksymalna siła - siła oporu

Siła = 1400 N - 400 N = 1000 N.

Stąd maksymalne przyspieszenie;

1000 N / 2000 kg

= 0,5 m / s²

Answer:

c i kn now it is

Explanation:

Answer:

a) La piedra es lanzada desde una altura de 0,785 metros.

b) La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Explanation:

a) Dado que la piedra es lanzada horizontalmente, tenemos que la piedra experimenta un movimiento horizontal a velocidad constante y uno vertical uniformemente acelerado debido a la gravedad. La altura de la que fue lanzada la piedra se puede determinar mediante la siguiente ecuación cinemática:

[tex]y = y_{o}+v_{o,y}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)

Donde:

[tex]y[/tex] - Altura final, medida en metros.

[tex]y_{o}[/tex] - Altura inicial, medida en metros.

[tex]v_{o,y}[/tex] - Componente vertical de la velocidad inicial, medida en metros por segundo.

[tex]t[/tex] - Tiempo, medido en segundos.

[tex]g[/tex] - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que [tex]y = 0\,m[/tex], [tex]v_{o,y} = 0\,\frac{m}{s}[/tex], [tex]t = 0,4\,s[/tex] y [tex]g = -9,807\,\frac{m}{s^{2}}[/tex], entonces la altura inicial de la piedra es:

[tex]y_{o} = y-v_{o,y}\cdot t -\frac{1}{2}\cdot g\cdot t^{2}[/tex]

[tex]y_{o} = 0\,m-\left(0\,\frac{m}{s} \right)\cdot (0,4\,s)-\frac{1}{2}\cdot \left(-9,807\,\frac{m}{s^{2}} \right) \cdot (0,4\,s)^{2}[/tex]

[tex]y_{o} = 0,785\,m[/tex]

La piedra es lanzada desde una altura de 0,785 metros.

b) Ahora, obtenemos el componente horizontal de la velocidad inicial a partir de la siguiente ecuación cinemática:

[tex]v_{o,x} = \frac{x-x_{o}}{t}[/tex] (2)

Donde:

[tex]x_{o}[/tex], [tex]x[/tex] - Posiciones horizontales iniciales y finales, medidas en metros.

[tex]t[/tex] - Tiempo, medido en segundos.

Si tenemos que [tex]x_{o} = 0\,m[/tex], [tex]x = 2,5\,m[/tex] y [tex]t = 0,4\,s[/tex], entonces el componente horizontal de la velocidad inicial es:

[tex]v_{o,x} = \frac{2,5\,m-0\,m}{0,4\,s}[/tex]

[tex]v_{o,x} = 6,25\,\frac{m}{s}[/tex]

La piedra es lanzada con una velocidad inicial de 6,25 metros por segundo.

Answer:

The net force acting on the wagon is 63 N

Explanation:

Net Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:

F = m.a

Where (a) is the acceleration of the object.

The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:

F = 42*1.5

F = 63 N

The net force acting on the wagon is 63 N

Answer:

The final velocity of the train and the moose after collision is approximately 19.51 m/s

Explanation:

The given mass of the moose, m₁ = 250 kg

The velocity of the moose, v₁ = 0

The mass of the oncoming train, m₂ = 10,000 kg

The velocity of the train, v₂ = 20 m/s

The velocity of the moose and the train after collision = v₃

By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision

m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃

Therefore, by substitution, we have;

250×0 + 10,000× 20 = (10,000 + 250) × v₃

200,000 = 10,250 × v₃

v₃ = 200,000/10,250 ≈ 19.51 m/s

The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s

Answer:

9.8×104Nm−2,0.025,3.92×106Nm−2

Solution :

Here, L=10cm=10×10−2m

F=100kgf=100×9.8N

ΔL=0.25cm=0.25×10−2m,

Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025

=3.92×106Nm−2

Explanation:

Answer:

Give me some time okkkk

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 = 50 × 0.6

We have the final answer as

Hope this helps you

Answer:

The net and only force acting on the ball thrown up into the air at half of its maximum height is the weight  of the ball

Explanation:

When the ball is rising in the air, the force, F, acting on it is given by the product of the mass, m × acceleration, a

The acceleration of a body thrown in the air = Gravitational acceleration = g = Constant

Therefore;

The force acting on the body thrown in the air F = Constant = m × g (downwards)

The force acting on the ball thrown up into the air at half of its maximum height = The mass of the ball × The acceleration due to gravity = The weight  of the ball.

Answer:

A subduction zone is also generated when two oceanic plates collide — the older plate is forced under the younger one — and it leads to the formation of chains of volcanic islands known as island arcs.

Explanation:

Answer:

It is B

Explanation:

Answer: 2nd answer

Explanation: took exam

Solution:

The magnitude of A vector is 84.9 m in the positive y-axis direction.

So the X component of A =0

     the Y component of A = 84.9 m

Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.

So  the X component of B = B cos 41°

                                           = 195 x cos 41°

                                           = 195 x 0.75   = 146.25 m

   the Y component of B = B sin 41°

                                           = 195 x sin 41°

                                           = 195 x 0.65   = 126.75 m

Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.

So the X component of C =126.2 m

     the Y component of C = 0

Comparing all these, we get

1. B vector has the largest X component

2. B vector has the largest Y component

Answer:

The effects of the Moon's rotation includes;

1) The Moon rotation and revolution gives the appearance of a perfectly still Moon to observers of the Moon from the Earth

2) The Moon has two sides, the near side that continuously faces the Earth and the "back" or far side, which is also known as the dark side of the Moon

The effect of the Moon's revolution

1) The tides in the ocean and water bodies, due to the gravitational forces from the Moon

2) The changes in the observed shape of the Moon due to the amount of Sunlight that is able to be reflected from the Moon as a result of the relative position of the Moon, the Earth and the Sun, at a given point in time

3) Lunar and Solar eclipse, when the Earth and the Moon blocks the light coming from the Sun respectively, due to their combined revolution

Explanation:

The duration of the Moon's orbit round the Earth = 27.322 days

The time it takes the Moon to rotate round its axis = 27 days

The Moon is the closest cosmic body to the Earth.

Answer:

yeet yeet yeet yeet

Explanation:

Kinetic energy (K.E):-

So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.

A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.

What is the squirrel’s mass?

Answer: 0.51 kg

Answer:

Explanation:

Power = Workdone/Time

Since workdone = Force * distance = mgh

Power = mgh/t

Given

Power = 10watts

speed = 2m/s

Power = Weight * velocity

Substitute

10 = Weight * 2

Weight = 10/2

Weight = 5N

Hence the weight of the object is 5N

The forces constant (spring constant) of the spring will be 41.09 N/m.

The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;

F = kx

The given data in the problem is;

F is the spring force = 30 N

K is the spring constant= ?

x is the displacement of spring = 0.73 m

The spring constant is;

K =F/x

K=30/0.73

K=41.09 N/m

Hence the force constant (spring constant) of the spring will be 41.09 N/m.

To learn more about the spring force refer to the link;

https://brainly.com/question/4291098

#SPJ1

Answer:mechanical waves.

Explanation:

Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.

Answer:

The water hits the wall at a height of 5.38 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]

Where:

y   = vertical displacement

x   = horizontal displacement

θ   = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

Answer:

33,333.33 N

Explanation:

Given that :

Initial kinetic energy = 500,000 J

Final kinetic energy = 100,000 J

Using the relation :

Force * time = change in momentum (Newton's law)

Force (F) * 0.12 = (500,000 - 100,000)

0.12F = 400,000 J

Force = (400,000 J) / 0.12s

Force = 33333.333

Force = 33,333.33 N

Answer: The work done on the flagstone is 20178 J

Explanation:

Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.

[tex]Work=Power\times time[/tex]

Given : work = ?

Power = 342 W = 342J/s

Time = 59 s

[tex]Work=342J/s\times 59s=20178J[/tex]

Thus the work done on the flagstone is 20178 J

Answer: Number 6 is Periods

Explanation:

Answer:

a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

b. -2.81 m/s²

c. 3.78 s

d. -351.25 N

Explanation:

a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.

For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.

The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².

For the block of ice, v = 15 m/s and Ф = 45°. So,

h = v²sin²Ф/2g

= (15 m/s)²sin²45/(2 × 9.8 m/s²)

= 225 (m/s)²(1/√2)²/19.6 m/s²

= 225 (m/s)²(1/2)/19.6 m/s²

= 112.5 (m/s)²/19.6 m/s²

= 5.74 m

Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.

The graph is in the attachment.

b. What is the rate of acceleration while the blocks slow to a stop?

Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m

So, a =  (v² - u²)/2s

substituting the variables into the equation, we have

a =  ((0 m/s)² - (10.61 m/s)²)/2(20 m)

= - 112.57 (m/s)²)/40 m

= -2.81 m/s²

c. How long do the blocks take to slow to a stop?

Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop

So, making t subject of the formula,

t = (v - u)/a

substituting the values of the variables, we have

t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²

= -10.61 m/s/-2.81 m/s²

= 3.78 s

d.  What is the amount of friction between the ice and the snowy ground?

The frictional force, f = net force on block of ice

f = ma where  m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²

f = ma

= 125 kg(2.81 m/s²)

= -351.25 N

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 15 × 2

We have the final answer as

Hope this helps you