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3 years ago

(03.05 LC)

Physics

2 answers:

Lynna [10]3 years ago

7 0

Honed I don’t know where the question is

Nikitich [7]3 years ago

4 0

I dont see the question to this

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PLEASE HELP ASAP WILL GIVE BRAINLIEST

stealth61 [152]

Explanation

(m) is measured in kilograms (kg)

<h2>(F) is measured in newtons (N)</h2>

<h3>acceleration (a) is measured in metres per second squared (m/s²)</h3>

4 0

3 years ago

Read 2 more answers

A boy on a bicycle drags a wagon full of newspapers at 0.80 m/s for 30 min (1800 s) using a force of 40 N. How much work has the

Eva8 [605]

Answer:

N133A

Explanation:black

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2 years ago

Sam is riding his bike in the park when he sees a soccer ball rolling toward him. He applies his brakes to slow down. Once he st

nlexa [21]

When Sam presses the brake lever, a pair of rubber shoes clamps onto the metal inner rim of the front and back wheels. As the brake shoes rub against the wheels, friction is caused and the kinetic energy possessed by the vehicle is converted into heat which slows down the vehicle.

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3 years ago

A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/

luda_lava [24]

Answer:

1) The magnitude of the angular acceleration = 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration = 2.744 m/ $s^{2}$

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed $v_{0}$ = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

$F_{g}$ = μ N , N = m g

$F_{g}$ = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m $R^{2}$

torque τ = I α

τ = F R

I α = F R

(2/5) m $R^{2}$ α = μ m g R

α = (5 μ g / 2R) μ g R

= (5 x 0.28 x 9.8/ 2 x 0.101)

= 67.92 rad/ $s^{2}$

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - $F_{g}$ , $F_{k}$ is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

= (0.28) (9.8)

= 2.744 m/ $s^{2}$

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, $v_{t}$ = $v_{0}$ - a t

$v_{t}$ = $v_{0}$ - μ g t

the angular speed, ω = ω0 + α t

ω = ω0 + (5 μ g/2R ) t

$v_{t}$ = ω R

$v_{0}$ - μ g t = ω0 R + (5 μ g/2R ) t R

7 μ g t/2 = $v_{0}$ + ω0 R

hence,

t = (2 $v_{0}$ + ω0 R)/ 7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 $v_{0}$ / 7 μ g

= 2 x 8.7 / 7 (0.28) (9.8)

= 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = $v_{0}$ t - (1/2) a $t^{2}$

= (8.7) (0.906) - (1/2) (2.744) $0.906^{2}$

= 6.75 m

5) The final velocity

$v_{t}$ = $v_{0}$ - a t

$v_{t}$ = 8.7 - (2.744) (0.906)

$v_{t}$ = 6.21 m/s

4 0

3 years ago

Why is evidence of meteorite impact rare on earth, despite the fact that meteoroids often pass into earth's atmosphere?

DENIUS [597]

The fact is your rooftop is covered with dust particles from meteorites. Outerspace is full of debris. Every day, thousands of meteorites invade our home but thanks to atmosphere for keeping us safe from them :D
When these meteorites enter our atmosphere, they experience resistance to the particles present in our atmosphere. As a result they catch fire(and eventually melts) and lose all their energy before falling on the surface. This happens with very small sized meteorites but when bigger ones enter, they surely hit the surface because they are difficult to melt.
Thanks to Jupiter for saving us from asteroids. Bigger ones rarely enters our atmosphere because of Jupiter's gravitational pull. It lures all the debris towards itself (:

5 0

3 years ago