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Dominik [7]
3 years ago
15

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

cceleration?
2.0 m/s/s
2.5 m/s/s
5.0 m/s/s
20. m/s/s
50. m/s/s
Physics
2 answers:
vladimir1956 [14]3 years ago
8 0
I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s
Genrish500 [490]3 years ago
4 0

Answer:

a = 20 m/s²

Explanation:

The magnitude of the acceleration of an object that moves in a circular path is given by:

a= \sqrt{(a_{t})^{2} +(a_{n})^{2} } Formula (1)

where:  

at = α*R : Formula (2)  :Tangential acceleration

ac = ω²*R :  Formula (3) : Centripetal acceleration

α : angular acceleration (rad/s²)  

ω: angular speed  (rad/s)  

R : is radius where the object is located from the center of the circular path

The tangential velocity of the body is calculated as follows:

v = ω*R  Formula (4)

where:  

v is the tangential velocity or linear velocity (m /s)  

ω is the angular speed (rad/s)  

R is radius where the body is located from the center of the circular path

Data

v = 10 m/s : tangential speed of the object (uniform)

R = 10 m

Calculating of ω  (angular speed)

We replace data in the formula (4)

v = ω*R

10 = ω*5

ω = 10 / 5

ω = 2 rad/s

Calculating of the Centripetal acceleration (ac)

We replace ω = 2 rad/s in the formula (3)

ac = ω²*R

ac = (2)²*(5)

ac = 20 m/s²

Calculating of  the tangential acceleration (at)

Because the magnitude of the tangential speed is uniform , then α=0

We replace α=0  in the formula (2)

at =α*R = 0*5

at = 0

Calculating of  the magnitude of the object's acceleration (a)

We replace :  at = 0 and ac = 20 m/s²  in the formula (1)

a= \sqrt{(0)^{2} +(20)^{2} }

a = 20 m/s²

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Which pair of countries are both part of the crib Caribbean cultural regions in North America
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3 years ago
Automobile A and B are initially 30 m apart travelling in adjacent highway lanes at speeds VA = 14.4 km/hr., VB 23.4 km/hr. at t
marshall27 [118]

Answer:

        x = 240 m

Explanation:

This is a kinematics exercise

Let's fix our frame of reference on car A

           x = x₀ₐ+ v₀ₐ t + ½ aₐ t²

         

the initial position of car a is zero

           x = 0 + v₀ₐ t + ½ 0.8 t²

for car B

          x = x_{ob} + v_{ob} t - ½ a_b t²

     

car B's starting position is 30 m

         x = 30 + v_{ob} t - ½ 0.4 t²

at the point where they meet, the position of the two vehicles is the same

         0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²

let's reduce the speeds to the SI system

        v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s

        v_{ob} = 23.4 km / h = 6.5 m / s

        4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²

        0.2 t² - 2.5 t - 30 = 0

        t² - 12.5 t - 150 = 0

we solve the quadratic equation

       t = \frac{12.5 \pm \sqrt{12.5^2 + 4 \ 150}  }{2}

       t = \frac{12.5 \  \pm 27.5}{2}

       t₁ = 20 s

       t₂ = -7.5 s

time must be a positive quantity so the correct result is t = 20 s

let's look for the distance

        x = 4 t + ½ 0.8 t²

        x = 4 20 + ½ 0.8 20²

        x = 240 m

8 0
3 years ago
70.5
ra1l [238]

Answer:5000000000hertz

Explanation:

Wavelength=6cm=6/100 m=0.06m

Frequency=velocity/wavelength

Frequency=(3×10^8)÷0.06

Frequency=5000000000 hertz

5 0
3 years ago
Helppppp plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz!
Fynjy0 [20]
C real,inverted and smaller than the object
6 0
3 years ago
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