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Charra [1.4K]
3 years ago
14

A robot standing on a cliff shoots a ball upwards with an initial speed of 30 m/s. What is the height of the cliff given that th

e ball reaches the bottom of the cliff 8 s after the shoot? (Take g = 10 m/s^2 and the height of the robot is negligible.)
A 25 m
B 45 m
C 80 m
D 145 m​

Physics
1 answer:
valina [46]3 years ago
5 0

Answer:

C 80 m

Explanation:

Given:

v₀ = 30 m/s

a = -10 m/s²

t = 8 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (30 m/s) (8 s) + ½ (-10 m/s²) (8 s)²

Δy = -80 m

The ball lands 80 m below where it started.  So the height of the cliff is 80 m.

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The high-pressure air storage tank for a supersonic wind tunnel has a volume of 1000 ft3. If air is stored at a pressure of 30 A
Yuri [45]

Answer:

69.86 slugs

Explanation:

From the ideal gas equation

PV = nRT

n (number of moles) = PV/RT

P is pressure of the gas = 30 atm = 30×101325 = 3,039,750 Pa

V is volume of the gas = 1000 ft^3 = 1000/3.2808^3 = 28.32 m^3

R is gas constant = 8.314 J/mol.K

T is the temperature of the gas = 530 R = 5/9 × 530 = 294.44 K

n = 3,039,750×28.32/8.314×294.44 = 35166.11 mol

mass = n× MW

MW of air = 29 g/mol

mass = 35166.11 × 29 = 1019817.19 g × 0.0685 slug/1000 g = 69.86 slugs

5 0
4 years ago
What is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?
Art [367]

CORRECT ANSWER:

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

STEP-BY-STEP EXPLANATION:

The complete question from book is

According to Figure 9.6, what is a key difference between cell signaling by a cell-surface receptor and cell signaling by an intracellular receptor?

a- Cell-surface receptors bind polar signaling molecules; intracellular receptors bind nonpolar signaling molecules.

b- Signaling molecules that bind to cell-surface receptors lead to cellular responses restricted to the cytoplasm; signaling molecules that bind to intracellular receptors lead to cellular responses restricted to the nucleus.

c- Cell-surface receptors bind to specific signaling molecules; intracellular receptors bind any signaling molecule.

d- Cell-surface receptors typically bind to signaling molecules that are smaller than those bound by intracellular receptors.

e- None of the other answer options is correct.

3 0
3 years ago
A flat metal washer is heated. As the washer's temperature increases, what happens to the hole in the center? A flat metal washe
STALIN [3.7K]

To solve this exercise it is necessary to take into account the concepts related to thermal expansion.

The thermal expansion is given by the function,

\Delta L = L_0 \alpha \Delta T

Where,

\Delta L= Change in Length

\Delta T =Change in Temperature

\alpha =Coeficiente de dilatación lineal

L_0 = Initial Length

By quickly deducing the formula, we can realize that the greater the change in temperature, the greater the change in the length of the radius.

The change in length is proportional to the change in temperature. Considering that the other two terms are constant we have that the correct one would be: <em>The hole in the center of the washer will expand.</em>

4 0
3 years ago
Two parallel plates that are initially uncharged are separated by 1.7 mm, have only air between them, and each have surface area
yaroslaw [1]

Answer:

5.63\cdot 10^{-6} C

Explanation:

The capacitor of a parallel-plate capacitor is given by:

C=\epsilon_0 \frac{A}{d}

where

A is the area of each plate

d is the separation between the plates

\epsilon_0 is the vacuum permittivity

The energy stored in a capacitor instead is given by

U=\frac{1}{2}\frac{Q^2}{C}

where

Q is the charge stored in each plate

Substituting the expression we found for C inside the last formula,

U=\frac{1}{2}\frac{Q^2 d}{\epsilon_0 A}

And re-arranging it

Q=\sqrt{\frac{2U\epsilon_0 A}{d}}

Now if we substitute

d=1.7 mm=0.0017 m\\A=16 cm^2 = 16\cdot 10^{-4} m^2\\U = 1.9 J

We find the charge stored on the capacitor:

Q=\sqrt{\frac{2(1.9)(8.85\cdot 10^{-12})(16\cdot 10^{-4})}{0.0017}}=5.63\cdot 10^{-6} C

7 0
3 years ago
What do you think we can learn about Earth's history by looking at the Grand Canyon?
viktelen [127]
We can look at all the ages of the earth since it’s a big crack is reveals many layers of the earth and we can know about chemicals and metals that were in earth and diffrent times
5 0
3 years ago
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