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7nadin3 [17]
3 years ago
11

First, you will investigate purely vertical motion. The kinematics equation for vertical motion (ignoring air resistance) is giv

en byy(t)=y0+v0t−(1/2)gt2,where y0=0 is the initial position, v0 is the initial speed, and g is the acceleration due to gravity.Drag the cannon downwards so it is at ground level, or 0 m (which represents the initial height of the object), then fire the pumpkin straight upward (at an angle of 90∘) with an initial speed of 14 m/s.How long does it take for the pumpkin to hit the ground?
Physics
2 answers:
Ivenika [448]3 years ago
8 0

Answer:

2.85 s .

Explanation:

y(t) = y(0) + v₀t + 1/2 gt²

y(t) is vertical displacement , y(0) is initial position , v₀ is initial velocity and t is time required to make vertical displacement and g is acceleration due to gravity.

Here  y(0) is zero , v₀ = 14 m/s , g = 9.8 m s⁻² , y(t ) = 0 , as the pumpkin after time t comes back to its initial position, that is ground .

We shall take v₀ as negative as it is in upward direction and g as positive as it acts in downward direction

Put the values in the equation above,

0 = 0 - 14t + 1/2 x 9.8 t²

14 t = 1/2 x 9.8 t²

t = 28 / 9.8

t = 2.85 s .

mamaluj [8]3 years ago
3 0

Answer:

2.85 s .

Explanation:

y(t) = y(0) + v₀t + 1/2 gt²

y(t) is vertical displacement , y(0) is initial position , v₀ is initial velocity and t is time required to make vertical displacement and g is acceleration due to gravity.

Here  y(0) is zero , v₀ = 14 m/s , g = 9.8 m s⁻² , y(t ) = 0 , as the pumpkin after time t comes back to its initial position, that is ground .

We shall take v₀ as negative as it is in upward direction and g as positive as it acts in downward direction

Put the values in the equation above,

0 = 0 - 14t + 1/2 x 9.8 t²

14 t = 1/2 x 9.8 t²

t = 28 / 9.8

t = 2.85 s .

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A solid brass cylinder and a solid wood cylinder have the same radius and mass (the wood cylinder is longer). Released together
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Explanation:

a.

The both cylinders will reach the bottom at the same time notice the relation in the equation in indepent of the length and both have the same radius and the same rotational inertia.

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a=\frac{g*sin(\beta)}{1+I_{com}/m*r^2}

So both will be tie

b.

a_{brass}=\frac{g*sin(\beta)}{1+I_{brass}/m*r^2}=a_{wood}=\frac{g*sin(\beta)}{1+I_{wood}/m*r^2}

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3 years ago
A charge q1 of -5.00 x 10^-9 C and a charge q2 of -2.00x 10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium po
aleksandrvk [35]

Answer:

Explanation:

Let the equilibrium position of third charge be x distance from q₁.

Force on third charge due to q₁

= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²

Force on third charge due to q₂

= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

Both the force will act in opposite direction and for balancing , they should be equal.

9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²

5  / x² = 2 / ( .4 - x )²

Taking square root on both sides

2.236 / x = 1.414 / .4 - x

2.236 ( .4 - x ) = 1.414 x

.8944 - 2.236 x = 1.414 x

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So the third charge should be at a distance of 24.5 cm from q₁ .

4 0
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Answer:

0.3659

Explanation:

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since both the bulbs radiate same power

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Now e₁=e₂

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or

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or

\frac{A_1}{A_2} }=0.3659

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