answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tanya
12 days ago
13

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s

ame location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward. S is constant because at the moment Rock Y is released, the only difference between the rocks is their difference in height above the ground. (B) S is constant because the difference in speed between the two rocks stays constant as they fall C S increases because the difference in speed between the two rocks increases as they fall S increases because at all times Rock X falls with a greater speed than Rock Y
Physics
1 answer:
Yuliya22 [1.1K]12 days ago
3 0

Answer:

C) True. The distance S increases over time, with v₁ = gt and v₂ = g (t-t₀), illustrating that v₁> v₂ for the same t.

Explanation:

We have a set of statements to evaluate for correctness. The most effective approach is to examine the problem in detail.

Using the equation for vertical launch, we acknowledge that the positive direction signifies downward movement.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Stone 2

Released shortly thereafter, let's assume a delay of one second, we can utilize the same timing mechanism

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

We can now calculate the separation distance between the two stones, which is applicable for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t² - 2 t t₀ + t₀²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t - t₀)

This represents the distance between the two stones over time, with the coefficient outside the parentheses being constant.

For t < to, the first stone remains stationary while the distance grows.

For t > = to, the expression (2t/to-1) yields a value greater than 1, indicating that the distance expands as time progresses.

We can now analyze the different statements

A) false. The height difference increases over time.

B) False S increases.

C) It is true that S increases over time, with v₁ = gt and V₂ = g (t-t₀) indicating v₁> v₂ at the same t.

You might be interested in
Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller
Softa [913]

Answer:

σ₁ = 3.167 * 10^{-6} C/m²

σ₂ = 7.6 * 10 ^{-6}  C/m²

Explanation:

Provided Information:

i) Smaller sphere's radius ( r ) = 5 cm.

ii) Larger sphere's radius ( R ) = 12 cm.

iii) Electric field at the larger sphere's surface  ( E₁ ) = 358 kV/m, which is equivalent to 358 * 1000 v/m

E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )

358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }

Charge (Q₁) = 572.8 * 10^{-9} C

Since the electric field inside a conductor is zero, the electric potential ( V ) remains constant.

V = constant

∴\frac{Q_{1} }{R} = \frac{Q_{2} }{r}

Q_{2} = \frac{r}{R} *Q_{1}

Q_{2} = \frac{5}{12} *572.8*10^{-9}   = 238.666 *10^{-9} C

Surface charge density ( σ₁ ) for the larger sphere.

Calculated Area ( A₁ )  = 4 * π * R²  = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁  = \frac{Q_{1} }{A_{1} } = \frac{572.8 *10^{-9} }{0.180864} = 3.167 * 10^{-6}  C/m².

Surface charge density ( σ₂ ) for the smaller sphere.

Calculated Area ( A₂ )  = 4 * π * r²  = 4 * 3.14 * 0.05²  = 0.0314 m².

σ₂ =\frac{Q_{2} }{A_{2} } = \frac{238.66 *10^{-9} }{0.0314} = 7.6 * 10 ^{-6} C/m²

8 0
4 days ago
a fixed mass of a n ideal gas is heated from 50 to 80C at a constant pressure at 1 atm and again at a constant pressure of 3 atm
inna [987]

Answer:

The required energy remains identical in both scenarios since the specific heat capacity (Cp) does not change with varying pressure.

Explanation:

Given;

initial temperature, t₁ = 50 °C

final temperature, t₂ = 80 °C

Temperature change, ΔT = 80 °C - 50 °C = 30 °C

Pressure for scenario one = 1 atm

Pressure for scenario two = 3 atm

The energy needed in both scenarios is expressed as;

Q = M*C_p*\delta T

Where;

Cp denotes specific heat capacity, which only varies with temperature and remains unaffected by pressure.

Hence, the energy required remains the same for both scenarios since specific heat capacity (Cp) is pressure-independent.

8 0
9 days ago
What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?
Yuliya22 [1153]
In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

6 0
22 hours ago
When 999mm is added to 100m ______ is the result​
Keith_Richards [1034]

Answer:

The outcome of adding 999mm to 100m is 101m.

Explanation:

That's my belief.

6 0
5 days ago
Read 2 more answers
To determine the height of a flagpole, Abby throws a ball straight up and times it. She sees that the ball goes by the top of th
Softa [913]

Answer:

H = 10.05 m

Explanation:

The stone reaches the top of the flagpole at both t = 0.5 s and t = 4.1 s

therefore, the total duration of the upwards motion above the peak of the pole is provided as

\Delta t = 4.1 - 0.5 = 3.6 s

now we have

\Delta t = \frac{2v}{g}

3.6 = \frac{2v}{9.8}

v = 17.64 m/s

this indicates the speed at the flagpole's top

at this point we have

v_f - v_i = at

17.64 - v_i = (-9.8)(0.5)

v_i = 22.5 m/s

the height of the flagpole is stated as

H = \frac{v_f + v_i}{2}t

H = \frac{22.5 + 17.64}{2} (0.5)

H = 10.05 m

5 0
12 days ago
Other questions:
  • f a stadium pays $11,000 for labor and $7,000 for parking, what would the stadium's parking revenue have to be if the stadium is
    5·2 answers
  • Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
    15·2 answers
  • A clever inventor has created a device that can launch water balloons with an initial speed of 85.0 m/s. Her goal is to pass a b
    8·1 answer
  • A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass
    8·1 answer
  • A running mountain lion can make a leap 10.0 mlong, reaching a maximum height of 3.0 m. What is the speed of the mountain lion j
    10·1 answer
  • A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W
    7·1 answer
  • Phyllis had always been the one to take care of Douglas, now he feels as if it is his duty and privilege to take care of her, ev
    13·2 answers
  • A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle
    13·1 answer
  • Energy is observed in two basic forms: potential and kinetic. Which of the following correctly matches these forms with a source
    7·1 answer
  • An Object moving at a velocity of 30 m/s slows to a stop in 7 seconds. What was its acceleration
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!