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.
If we connect Gabe's current position to the top of the building, to the bottom of the building and back to Gabe, we form a right triangle with the height of 250 m and the angle opposite to the height being equal to 19°. If we let x be his distance from the building, we use the trigonometric formula,
tan 19° = x/250 m
The value of x from the equation is 86.08 m. Thus, the answer is 86.08 m.
The magnitude of v⃗ is {sqrt(m1v1)2+(m2v2)2/ m1+m2}, that is, the speed v of the two-car unit after the collision.
<h3>What is
collision?</h3>
- In physics, collisions occur when particles, aggregates of particles, or solids come close to each other, interact and affect each other.
- Collisions are of three types Fully elastic collision, inelastic collision and Perfectly inelastic collision.
- Multiply the mass of the second object by its velocity.
- For example, if the weight is 1,000 and the speed is -30 meters per second, then its momentum is 30,000 kg meters per second.
- Add the two velocities together to determine the direction the object will move after a collision.
- So the formula for determining the size of a vector (in 2D space) is v = (x, y).|v| = √(x2 + y2).
- This formula is derived from the Pythagorean theorem.
- The formula V = (x, y, z) that determines the size of a vector (in 3-dimensional space) is:|V| = √(x2 + y2 + z2)
To learn more about collision from the given link :
brainly.com/question/13138178
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B strength training I think that’s the answer
Concept:
Frequency- It is defined as the number of oscillations occur in one second.
Its SI unit is Hertz (Hz)
Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds
∵ In 0.75 second, produced sound has oscillations = 18,500 cycles
∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz
The frequency of the sound will be ≈ 24,667 Hz
From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.