Why do scientist use models to study atoms

A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C . Dry steam at 100°C is passed in unt

Lina20 [59]

Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

7 0

3 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

v = v₀ - a t

a = (v₀- v) / T

On planet X

v = v₀ - a' t’

a ’= (v₀-v) / 2T

Let's substitute the land values in plot X

a’= a / 2

Now let's use Newton's second law

W = ma

m g = m a

a = g

We substitute

a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0

3 years ago

A projectile lands at the same height from which it was launched. which initial velocity will result

Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the air resistance is ignored, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

5 0

3 years ago

Suppose you exert 150 n on your refrigerator and push it across the kitchen floor at constant velocity. what friction force acts

Pachacha [2.7K]

Due that the velocity is constant that means that friction force is equal to the force exert by you, otherwise the refrigerator will accelerate or decelerate and in both cases velocity will not be constant.
So then the friction force between refrigerator and floor is 150 Newtons.

8 0

3 years ago

Read 2 more answers

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s