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3 years ago

A boy and his dog are running. The dog has a mass of 20kg and the boy has a mass of 45kg.

Physics

1 answer:

Ad libitum [116K]3 years ago

4 0

Answer:

Explanation:

mass of boy =m1=45 kg

mass of dog =m2=20kg

the speed of boy=3 m/s

the speed of dog =3 m/s

K.E of boy

K.E=1/2mv²

K.E=1/2(45)(3)²

K.E=202.5 J

K.E of dog

K.E=1/2mv²

K.E=1/2(20)(3)²

K.E=90 J

now the new speed of dog is 6 m/s

therefore new K.E of dog is

K.E=1/2(20)(6)²

K.E=360 j

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3 years ago

compare 51.5 hectograms to 51500 decigrams. How do each of these values compare to a gram, and which represents a larger mass?

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3 years ago

At the presentation ceremony, a championship bowler is presented a 1.64-kg trophy which he holds at arm's length, a distance of

solong [7]

To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.

By definition the torque is given by,

$\tau = F*d$

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$\tau = F*d$

For Newton's second law

$\tau = mg*d$

$\tau = 1.64*9.81*0.655$

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4 years ago

Five soccer players argued about when a soccer ball has energy. This is what they said.

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3 years ago

If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun

zvonat [6]

Answer:

H = 1/2 g t^2 where t is time to fall a height H

H = 1/8 g T^2 where T is total time in air (2 t = T)

R = V T cos θ horizontal range

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cos θ = 3 g T / (4 V) (I)

Now t = V sin θ / g time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V) from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97 6 within rounding

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3 years ago