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3 years ago

Examples of angular motion

Physics

2 answers:

MakcuM [25]3 years ago

6 0

Answer:

the swing of a baseball bat

the leverage on a hokey stick

virtually any club,stick,bat,racket that is swung

a runner on a circular path

ohaa [14]3 years ago

3 0

Answer:

A figure skater doing a double axle

The swing of a baseball bat

The leverage on a hockey stick

hope it helps

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Please I really need help on this please help thank you

maw [93]

What do you need help with?

4 0

3 years ago

Salmon often jump waterfalls to reach their

Y_Kistochka [10]

Answer:

5.38 m/s

Explanation:

Given (in the x direction):

Δx = 2.45 m

v₀ = v cos 42.5°

a = 0 m/s²

Δx = v₀ t + ½ at²

(2.45 m) = (v cos 42.5°) t + ½ (0 m/s²) t²

2.45 = (v cos 42.5°) t

t = 3.32 / v

Given (in the y direction):

Δy = 0.373 m

v₀ = v sin 42.5°

a = -9.8 m/s²

Δx = v₀ t + ½ at²

(0.373 m) = (v sin 42.5°) t + ½ (-9.81 m/s²) t²

0.373 = (v sin 42.5°) t − 4.905 t²

0.373 = (v sin 42.5°) (3.32 / v) − 4.905 (3.32 / v)²

0.373 = 2.25 − 54.2 / v²

v = 5.38

Graph:

desmos.com/calculator/5n30oxqmuu

4 0

3 years ago

A star near the visible edge of a galaxy travels in a uniform circular orbit. It is 41,200 ly (light-years) from the galactic ce

Ronch [10]

The total mass of the galaxy is 443.4 Solar mass

Orbital velocity ( $v$ ) = $\sqrt{\frac{MG}{R} }$

where M= weight of galaxy

G= gravitational constatnt = $6.674*10^-^1^1$ (given)

R = distance from centre = $41200$ Light years (given)= $4.12*9.5*10^1^6$ km (1 ly= $9.5*10^3$ billion km)

v= orbital velocity = $275$ $km/s$ (given)

∴ According to the formula

$(2.75*10^2)^2$ = $\frac{M*6.674*10^-^1^1}{4.12*9.5*10^1^6}$

⇒ $7.56*10^4*4.12*9.5*10^1^6=M*6.674*10^-^1^1$ (cross multiplying and expanding)

⇒ $29.59*10^2^1=M*6.674*10^-^1^1$

⇒ $\frac{29.59*10^2^1*10^1^1}{6.674}=M$

⇒ $4.434*10^3^2=M$

1 solar mass = $1.989*10^3^0 kg$

⇒ Mass in solar mass =443.4 Solar mass

⇒ M = 443.4 Solar mass

Learn more about Orbital velocity here :

brainly.com/question/22247460

#SPJ10

6 0

2 years ago

A 15 kg uniform disk of radius R = 0.25 m has a string wrapped around it, and a m = 3.4 kg weight is hanging on the string. The

Artyom0805 [142]

Answer:

5.45 J

Explanation:

When the 3.4 kg weight is moving with a speed of 1.1 m/s, what is the kinetic energy of the entire system?

RKE = $\frac{1}{2}I \omega ^2$

where;

$I = \frac{1}{2} mr^2$

$I = \frac{1}{2}*15*0.25^2$

$I = 0.46875 kg.m^2$

$\omega = \frac{1.0}{0.25}$

$\omega = 4 rad/s$

RKE = $\frac{1}{2}I \omega ^2$

= $\frac{1}{2} *0.46875*4^2$

= 3.75 J

LKE = $\frac{1}{2} mv^2$

= $\frac{1}{2} *3.4*1.0^2$

= 1. 7 J

K.E = RKE + LKE

K. E = ( 3.75 + 1.7 ) J

K . E = 5.45 J

3 0

3 years ago

Read 2 more answers

I just need to know if I am correct. thanks!

Iteru [2.4K]

Yes I believe you’re right but if not I’m sorry because I would’ve picked that answer as well lol

3 0

3 years ago

Examples of angular motion​

Examples of angular motion