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Jlenok [28]
4 years ago
10

A stone is thrown vertically upward with velocity 9.8ms 1)calculate the time taken to reach maximum height.2)what is the maximum

height reached 3) what is time taken to reach ground 4)what velocity it reaches the ground
Physics
2 answers:
agasfer [191]4 years ago
6 0

Answer:

It would take 6 seconds before hitting the ground

Explanation:

ira [324]4 years ago
5 0

Answer:

1.t=-1.96sec

2.H=4.8m

3.T=1.96sec

4.R=19.2m

Explanation:

u=9.8,t=?,sin theta=1

using formula t=2usintheta/g

t=2x9.8x1=19.6/10

t=1.96seconds

using formula H=u(squared)sin(squared)theta/2g

H=9.8(squared)x1(squared)/2x10

H=96x1/20

H=96/20

H=4.8m

using formula T=2usintheta/g

T=2x9.8x1/10

T=19.6/10

T=1.96sec

using the formula R=u(squared)sin2theta/g

R=9.8(squared)x2/10

R=96x2/10

R=192.08/10

R=19.2

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3 years ago
A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 6.5 μg is floating a dista
fenix001 [56]

Answer:

Q₂ = (9.83 × 10⁻⁹) C = +9.83 nC

Explanation:

The force of attraction/repulsion between the two charges is equal to the force of gravity on the charge 2.

The force of gravity on the charge two = mg

= 6.5 μg × 9.8 m/s² = (6.37 × 10⁻⁵) N

Since the gravity force is directed downwards, the force on charge 2 due to charge 1 has to be directed upwards, hence it is a force of repulsion.

The magnitude of the force between the two charges is given according to the Coulomb's law.

F = kQ₁Q₂/r²

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

Q₁ = 45 nC = (45 × 10⁻⁹) C

Q₂ = ?

r = 25 cm = 0.25 m

F = (6.37 × 10⁻⁵) N

(6.37 × 10⁻⁵) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] ÷ 0.25²

Q₂ = 0.00000000983 C = (9.83 × 10⁻⁹) C = 9.83 nC

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7 0
4 years ago
A playground ride consists of a disk of mass M = 49 kg and radius R = 1.7 m mounted on a low-friction axle. A child of mass m =
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Answer:

The angular speed is 0.83 rad/s.

Explanation:

Given that,

Mass of disk M=49 kg

Radius = 1.7 m

Mass of child m= 29 kg

Speed = 2.6 m/s

Suppose if the disk was initially at rest , now how fast is it rotating

We need to calculate the angular speed

Using conservation of momentum

m\omega_{i}=(mr^2+\dfrac{Mr^2}{2})\omega_{f}

mvR=(mr^2+\dfrac{Mr^2}{2})\omega

Put the value into the formula

29\times2.6\times1.7=(29\times1.7^2+\dfrac{49\times1.7^2}{2})\omega_{f}

\omega_{f}=\dfrac{29\times2.6\times1.7}{(29\times1.7^2+\dfrac{49\times1.7^2}{2})}

\omega_{f}=0.83\ rad/s

Hence, The angular speed is 0.83 rad/s.

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