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kompoz [17]
3 years ago
9

Understand that the acceleration vector is in the direction of the change of the velocity vector. In one dimensional (straight l

ine) motion, acceleration is accompanied by a change in speed, and the acceleration is always parallel (or antiparallel) to the velocity. When motion can occur in two dimensions (e.g. is confined to a tabletop but can lie anywhere in the x-y plane), the definition of acceleration is
Physics
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

 a = √ (a_t² + a_c²)

a_t = dv / dt ,    a_c = v² / r  

Explanation:

In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.

            a = aₓ i ^ + a_y j ^

Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains

         a = √ (a_t² + a_c²)

where the tangential acceleration is

           a_t = dv / dt

the centripetal acceleration is

          a_c = v² / r

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What is the kinetic energy of an object with a mass of 50kg and is and it is traveling at a rate of 60m/s.
liq [111]

Answer:

90,000 J

Explanation:

Kinetic energy can be found using the following formula.

KE=\frac{1}{2}mv^2

where <em>m </em>is the mass in kilograms and <em>v</em> is the velocity in m/s.

We know the object has a mass of 50 kilograms. We also know it is a traveling at a rate of 60 m/s. Velocity is the speed of something, so the velocity of the object is 60 m/s.

<em>m</em>=50

<em>v</em>=60

Substitute these values into the formula.

KE=\frac{1}{2}*50*60^2

First, evaluate the exponent: 60^2. 60^2 is the same as multiplying 60, 2 times.

60^2=60*60=3,600

KE=\frac{1}{2}*50*3,600

Multiply 50 and 3,600

KE=\frac{1}{2}*180,000

Multiply 1/2 and 3,600, or divide 3,600 by 2.

KE=90,000

Add appropriate units. Kinetic energy uses Joules, or J.

KE=90,000 Joules

The kinetic energy of the object is 90,000 Joules

6 0
3 years ago
Io experiences tidal heating primarily because __________. hints io experiences tidal heating primarily because __________. io i
maxonik [38]
Lo experiences tidal heating primarily because lo’s elliptical orbit causes the tidal force on lo to vary as it orbits the Jupiter. Thus, lo’s elliptical orbit is essential to its tidal heating. This elliptical orbit, in turn, is an end result of the orbital resonance among lo, Europa and ganymade. This orbital resonance origin lo to have a more elliptical orbit than it would because lo intermittently passes Europa and ganymade in the same orbital position. We cannot perceive tidal forces of tidal heating in lo but rather we foresee that they must occur based on the orbital characteristic of the moons and active volcanoes on lo is the observational evidence that tidal heating is significant in lo.
8 0
3 years ago
a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r
Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

f'=\frac{V+V_{o}}{V-V_{s}} f

Where:

f=8(10)^{4} Hz is the actual frequency of the sound wave

f'=8.002(10)^{4} Hz is the "observed" frequency

V=1570 m/s is the speed of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which are the red blood cells

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}

V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}

Finally:

V_{s}=0.392 m/s

3 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde
Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

  • The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal
  • The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

Volume \ of \ a \ cone = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot h

  • Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

  • \dfrac{dV}{dr} = \dfrac{d}{dr}  \left(\dfrac{1}{3} \cdot  \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot  s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0

3·r²·s² - 4·r⁴ = 0

3·r²·s² = 4·r⁴

3·s² = 4·r²

\underline{\left  \right. The \ radius, \, r =\sqrt{ \dfrac{3}{4}} \cdot s}

\underline{The \ height, \, h =\sqrt{s^2 - \dfrac{3}{4}\cdot s^2} = \dfrac{s}{2}}}

Learn more here:

brainly.com/question/14466080

3 0
3 years ago
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