Answer:
hii there
The correct answer is option ( C ) reduced air pressure
Explanation:
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Answer:
the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
Explanation:
Given the data in the question;
first we compute the total volume of mud
V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m
V = 2.508 × 10⁹ m³
same volume of mud spread uniformly over 1.4 × 1.4 km² area
Hence,
V = (1.4×1000)m × (1.4×1000)m × depth
2.508 × 10⁹ = 1400m × 1400m × depth
2.508 × 10⁹ m³ = 1960000m² × depth
depth = 2.508 × 10⁹ m³ / 1960000 m²
depth = 1279.59 m
so volume of the mud sitting above 3.5 m² area of the valley floor will be;
⇒ 3.5 m² × 1279.59 m
⇒ 4478.565 m³
so mass will be
m = 4478.565 × 1900 kg
m = 8509273.5 kg
Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
Answer:
NH₃
Explanation:
The compound that is covalent from the given choices is NH₃.
Covalent compounds are usually formed between two atoms with similar values of electronegativities such that the difference is very small or zero.
- This bond type involves the sharing of electrons between two atoms with similar electronegativities.
- Nitrogen and hydrogen forms stable configuration that are isoelectronic with noble gases by sharing their valence electrons.
- The 3 hydrogen electrons are enough to make nitrogen isoelectronic with neon.
- Also, the nitrogen with 3 lone pairs of electrons provides the bonding hydrogen with needed electrons to attain a structure similar to helium.
Explanation:
Simple Machine: Any of various devices that function in a manner basic to any machine, such as a lever, pulley, wedge, screw, or inclined plane. Complex Machine: A device consisting of two or more simple machines working together. ... A wedge is driven or forced between objects to split, lift, or make them stronger.
Answer:
.
Explanation:
The box is sliding with a constant speed in a fixed direction (to the right.) In other words the velocity of this box is constant. Hence, this box would be in a translational equilibrium. The acceleration of this box would be zero.
By Newton's Second Law of motion, the net force on this box would be . In other words, forces on this box are balanced.
The question is asking for the size of the friction on the box. Assuming that the floor is horizontal. The friction on this box would also be horizontal,
The only other force that could balance that friction would be the push to the right. The direction of this push is horizontal (to the right.) Hence, the entirety of that would be in the horizontal direction.
Thus, forces on this box in the horizontal direction would be:
- The push to the right.
- Friction that opposes the rightward motion of the box (that is, to the left.)
Since these two forces must balance each other, the size of the friction would also be .