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3 years ago

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A boat in a race accelerates uniformly at 0.2 m/s² for 18 seconds to reach a final speed of 6 m/s. Work out the initial speed of

the boat.

1 answer:

Xelga [282]3 years ago

4 0

Answer:

initial velocity=2.4m/s

Explanation:

Using the formula:

v=u+at

v= final velocity (6m/s)

u=initial velocity (???)

a=acceleration (0.2 m/s²)

t=time(18 seconds)

v=u+at

6=u+(0.2*18)

6=u+3.6

u=6-3.6

u (initial velocity)=2.4m/s

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Answer:

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A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/

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3 years ago

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A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

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3 years ago

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?

Daniel [21]

To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

$F=\frac{kq_{1}q_{2} }{r^{2} }$

Here, $q_{1}$ and $q_{2}$ are the charges on the pith balls, r is the separation between the charges and k is constant and its value is $8.99\times 10^9 N m^2/C^2$ .

Given $q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C$ and $r=8 cm=8\times10^{-2} m$ .

Substituting these values in above formula we get,

$F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9} N\\\\F=8.8\times10^{-4}N$

Thus, the repulsive force between two pith balls is $8.8\times10^{-4}N$ .

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3 years ago

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A stone is dropped from the top a 45m hign building how fast will it moving when it reaches the ground? Ande what will its veloc

posledela

Answer:

29.7 m/s fast, velocity is 29.7 m/s

Explanation:

Applying,

v² = u²+2gs...................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Given: u = 0 m/s (dropped from height), s = 45 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

v² = 0²+2×9.8×45

v² = 882

v = √(882)

v = 29.7 m/s.

Hence the stone will be moving 29.7 m/s fast and the velocity is also 29.7 m/s

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3 years ago