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solniwko [45]
3 years ago
15

A giant armadilo moving northward with a constant acceleration covers the distance between two points 60m apart in 6 seconds. It

s velocity as it passes the second point is 15m/s
what is the acceleration? (answer is 1.7) please explain how to solve this problem in detail.
Physics
1 answer:
Naddika [18.5K]3 years ago
8 0
Acceleration is the rate of change of the velocity of an object that is moving. This value is a result of all the forces that is acting on an object which is described by Newton's second law of motion. To determine acceleration, we need to know the initial velocity and the final velocity and the time elapsed. From the given values, we need t o calculate for the initial velocity. We use some kinematic equations. We do as follows:

 x = v0t + at^2/2
60 = v0(6) + a(6)^2/2
60 = 6v0 + 18a          (EQUATION 1)

vf = v0 + at
15 = v0 + a(6)
15 = v0 + 6a             (EQUATION 2)

Solving for v0 and a,
v0 = 5 m/s
a = 1.7 m/s^2
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What are some indicators of energy transformations?
Annette [7]

Explanation:

Contact, vision, sound, flavor, and smell are all markers of energy transformations. The most basic example would be when we notice something has begun to pass through vision. Whenever an entity accelerates or slows down, energy is constantly transformed.

5 0
2 years ago
Can anybody help me solve this problem? Thank you so much!
ser-zykov [4K]
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math

at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
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4 0
2 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
juin [17]

Answer:

Explanation:

Given

For first case

launch angle \theta =45^{\circ}C

at highest point h=150 m/s

150=u\cos 45

u=\frac{150}{\cos 45}=212.132 m/s

For second case

\theta _2=37^{\circ}C

at highest Point velocity is u\cos \theta _2

=212.132\times \cos 37

=169.41 m/s

as there is no acceleration in x direction therefore horizontal velocity is same          

7 0
3 years ago
Read 2 more answers
a telescope is oriting on a spacecraft aroudn the earth. Speed of 3.25 x 10^5 mass 7500 kg. What is the de Broglie wavelength of
Scrat [10]

Answer:

<em>2.72 x 10^-43 m</em>

<em></em>

Explanation:

mass of the telescope = 7500 kg

speed of the telescope = 3.25 x 10^5 m/s

de Broglie's  wavelength of the telescope is given as

λ = h/mv

where

λ is the wavelength of the telescope

h is the plank's constant = 6.63 × 10-34 m^2 kg/s

m is the mass of the telescope = 7500 kg

v is speed of the telescope = 3.25 x 10^5 m/s

substituting value, we have

λ = (6.63 × 10-34)/(7500 x 3.25 x 10^5)

λ = <em>2.72 x 10^-43 m</em>

8 0
3 years ago
An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s a
Cerrena [4.2K]

Assuming constant speeds, the P-wave covers a distance <em>d</em> in time <em>t</em> such that

9000 m/s = <em>d</em>/(60 <em>t</em>)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = <em>d</em>/(60(<em>t</em> + 1))

Now,

<em>d</em> = 540,000 <em>t</em>

<em>d</em> = 300,000(<em>t</em> + 1) = 300,000 <em>t</em> + 300,000

Solve for <em>t</em> in the first equation and substitute it into the second equation, then solve for <em>d</em> :

<em>t</em> = <em>d</em>/540,000

<em>d</em> = 300,000/540,000 <em>d</em> + 300,000

4/9 <em>d</em> = 300,000

<em>d</em> = 675,000

So the earthquake center is 675,000 m away from the seismic station.

6 0
3 years ago
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