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arlik [135]
3 years ago
6

A,B,C or D which of the following can be concluded regarding the movement represented in the graph?

Physics
1 answer:
Lerok [7]3 years ago
4 0

Answer:

C. It remained at a constant speed

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a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t
Zolol [24]

Answer: 0.146 m/s^{2}

Explanation:

The <u>centripetal acceleration</u> a_{c} of an object moving in a uniform circular path is given by the following equation:

a_{c}=\frac{V^{2}}{r}  (1)

Where:

V is the tangential velocity

r=1.05 m is the radius of the circle

On the other hand, the tangential velocity  is expressed as:

V=\omega r (2)

Where \omega is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

\omega=\frac{5 rev}{13.4 s}=0.37 rev/s (3)

Substituting (3) in (2):

V=(0.37 rev/s)(1.05 m) (4)

V=0.39 m/s (5)

Substituting (5) in (1):

a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}  (6)

Finally:

a_{c}=0.146 m/s^{2}  

6 0
3 years ago
Read 2 more answers
If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?
Bogdan [553]

It has 50kg with a velocity of 1 m/s times the speed of the cart divided by 2 and multiplied by kinectic x plus 5

3 0
3 years ago
Read 2 more answers
A bungee-jumping company operates on a bridge 200 m above the ground. They use bungee cords that are 100 m when they are unstret
zubka84 [21]

Answer:

m = 63.7 kg

Explanation:

As we know that when mass connected to the bungee cord stretch the string then the gravitational potential energy of the person will convert into potential energy of the string at the end

now here we know that when person jump from the top and reach at the end then loss in gravitational potential energy is given as

U = mgH

U = m(9.81)(200)

U = 1962 m

now when it is at the end of the motion stretch in the string will be

x = 200 - 100 = 100 m

now potential energy of string is given as

U_{spring} = \frac{1}{2}kx^2

U_{spring} = \frac{1}{2}(25)(100^2)

now by energy conservation we have

1962 m = \frac{1}{2}(25)(100^2)

m = 63.7 kg

6 0
3 years ago
N Rutherford's gold foil experiment the alpha particles pass through which part of the atom?
OleMash [197]
It passes through C the Electron Cloud
3 0
3 years ago
A merry-go-round rotates from rest with an angular acceleration of 1.04 rad/s2. How long does it take to rotate through (a) the
Andrei [34K]

Answer:

(a) 4.38 s.

(b) 1.817 s

Explanation:

(a)

Using

θ = ω₀t +1/2αt² ................ Equation 1

Where θ  = number of revolution, t = time, α = angular acceleration,  ω₀ = angular velocity.

Given: θ  = 1.59 rev = 1.59×2π = 9.992 rad,  ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

9.992 = 0(t) + 1/2(1.04)(t²)

t² = (2×9.992)/1.04

t² = 19.984/1.04

t = √(19.215)

t =4.38 s.

(b)

also using

θ = ω₀t +1/2αt²............... Equation 1

Given: θ  =3.18 rev = 3.18×2π = 19.97 rad,  ω₀ = 0 rad/s, α = 1.04 rad/s².

Substitute into equation 1

19.97 = 0(t) + 1/2(1.04)(t²)

t² = 19.97×2/1.04

t = √(38.40)

t = 6.197 s

The time require = 6.197-4.38 = 1.817 s

3 0
4 years ago
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