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3 years ago

16

Physics

2 answers:

frozen [14]3 years ago

7 0

Answer:

so you should try to find what things would cause the most risk. apply these to yourself as if you were doing this project and then think which two ways you can stay safe.

Explanation:

krek1111 [17]3 years ago

6 0

Explanation:

the table and the wooden block

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Un automovil con la velocidad de 5 m/s acelera durante 12 s a 3 m/s2,¿cual es la velocis

harina [27]

Answer:

60 miles

Explanation:

4 0

3 years ago

a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin

trapecia [35]

Answer:

a) $E_{p} = 0$

$E_{k} = 168.7 J$

$E_{m} = 168.7 J$

b) $E_{p} = 73.6 J$

$E_{k} = 95.8 J$

$E_{m} = 169.4 J$

c) $E_{p} = 169.2 J$

$E_{k} = 0$

$E_{m} = 169.2 J$

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

$E_{p} = mgh = 0$

The kinetic energy is:

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J$

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

$E_{p} = mgh = 1.5*9.81*5 = 73.6 J$

Now, to find the kinetic energy we need to calculate the speed at 5 m:

$v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9$

$v_{f} = \sqrt{126.9} = 11.3 m/s$

$E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J$

And the mechanical energies:

$E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J$

c) At its maximum height:

$v_{f}$ : is the final speed = 0

$h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m$

Now, the potential, kinetic and mechanical energies are:

$E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J$

$E_{k} = \frac{1}{2}mv^{2} = 0$

$E_{m} = 169.2 J + 0 = 169.2 J$

I hope it helps you!

7 0

3 years ago

A 160 g basketball has a 32.7 cm diameter and may be approximated as a thin spherical shell. Starting from rest, how long will i

mafiozo [28]

Answer:

t = 0.24 s

Explanation:

As seen in the attached diagram, we are going to use dynamics to resolve the problem, so we will be using the equations for the translation and the rotation dyamics:

Translation: ΣF = ma

Rotation: ΣM = Iα ; where α = angular acceleration

Because the angular acceleration is equal to the linear acceleration divided by the radius, the rotation equation also can be represented like:

ΣM = I(a/R)

Now we are going to resolve and combine these equations.

For translation: Fx - Ffr = ma

We know that Fx = mgSin27°, so we substitute:

(1) mgSin27° - Ffr = ma

For rotation: (Ffr)(R) = (2/3mR²)(a/R)

The radius cancel each other:

(2) Ffr = 2/3 ma

We substitute equation (2) in equation (1):

mgSin27° - 2/3 ma = ma

mgSin27° = ma + 2/3 ma

The mass gets cancelled:

gSin27° = 5/3 a

a = (3/5)(gSin27°)

a = (3/5)(9.8 m/s²(Sin27°))

a = 2.67 m/s²

If we assume that the acceleration is a constant we can use the next equation to find the velocity:

V = √2ad; where d = 0.327m

V = √2(2.67 m/s²)(0.327m)

V = 1.32 m/s

Because V = d/t

t = d/V

t = 0.327m/1.32 m/s

t = 0.24 s

7 0

3 years ago

A condition often referred to as being double jointed is an example of

Makovka662 [10]

It is an example of Hypermobility.

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3 years ago

1) Andrea and Chuck are riding on a merry-go-round. Andrea rides on a horse at the outer rim of the circular platform, twice as

telo118 [61]

Explanation:

The tangential speed of Andrea is given by :

$v=r\omega$

Where

r is radius of the circular path

ω is angular speed

The merry-go-round is rotating at a constant angular speed. Let the new distance from the center of the circular platform is r'

r' = 2r

New angular speed,

$v'=r'\omega'\\\\v'=(2r)\omega\\\\v'=2r\omega\\\\v'=2v$

New angular speed is twice that of the Chuck's speed.

8 0

3 years ago