Which of the following statements is an example of weather?

A set of twins works in the willis tower, a very tall office building in chicago. one works on the top floor and the other works

emmasim [6.3K]

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3 0

3 years ago

19

djyliett [7]

Answer:

B 23m/s

Explanation:

From the above information we get,

Initial velocity= 3m/s

Acceleration by gravity = 10 m/s2 (approx)

Time taken =2 seconds

v=u+at (First Equation Of Motion)

v= 3 + 10 × 2

v= 23 m/s

5 0

3 years ago

A rope exerts a force F on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of t

tia_tia [17]

Answer:

400 J

Explanation:

Given:

Δy = 4.00 m

v₀ = 0 m/s

a = 5.00 m/s²

Find: v²

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (5.00 m/s²) (4.00 m)

v² = 40.0 m²/s²

Find KE:

KE = ½ mv²

KE = ½ (20.0 kg) (40.0 m²/s²)

KE = 400 J

5 0

3 years ago

4. A massless spring hangs from the ceiling, and a mass is hung from the bottom of it. The mass is supported

MrRissso [65]

Answer:

The correct option is C: 0.31 s.

Explanation:

When the mass is then suddenly released we have:

$F = k\Delta y$

Where:

F is the force

k: is the spring constant

Δy: is the spring displacement

Since the tension in the spring is zero, the force is the weight:

$F = mg$

Where:

m is the mass of the object

g is the gravity

$mg = k\Delta y$ (1)

The oscillation period of the spring is given by:

$T = 2\pi \sqrt{\frac{m}{k}}$ (2)

By solving equation (1) for "k" and entering into equation (2) we have:

$T = 2\pi \sqrt{\frac{m}{\frac{mg}{\Delta y}}}$

$T = 2\pi \sqrt{\frac{\Delta y}{g}}$

Since the spring will osclliates in a position between the initial position (when it is at rest) and the final position (when the mass is released and reaches the bottom), we have Δy = 2.5 cm = 0.025 m:

$T = 2\pi \sqrt{\frac{0.025 m}{10 m/s^{2}}} = 0.31 s$

Hence, the oscillation period is 0.31 s.

The correct option is C: 0.31s.

I hope it helps you!

7 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago