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3 years ago

An archer fires an arrow at an angle of 9° above the horizontal with a resultant velocity of 24 m/s

Physics

1 answer:

Natali [406]3 years ago

5 0

Answer:

Explanation:

Here is the complete question. An archer fires an arrow at an angle of 9° above the horizontal with a resultant velocity of 24 m/s, How high will it go? What horizontal distance will it travel?

To determine the maximum height it will go, we will use the formula for calculating the maximum height as shown;

H = u²sin²theta/2g

u is the velocity = 24m/s

theta is angle of launch = 9°

g is acceleration due to gravity = 9.8m/s²

Substitute

H = 24²(sin9°)²/2(9.8)

H = 576×0.0245/19.6

H = 14.0957/19.6

H = 0.719m

Hence the arrow can travel 0.719m High

The horizontal distance it will travel is known as RANGE. This is expressed as;

R = U²sin2(theta)/g

R = 24²sin18°/9.8

R = 576sin18°/9.8

R = 177.99/9.8

R = 18.16m

Hence the horizontal distance the arrow can travel is 18.16m

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The figure above shows the net force exerted on an object as a function of the position of the object. The object starts from re

weqwewe [10]

Answer:

0.06 Kg

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Net Force (F) = 3 N

Mass (m) =?

Next, we shall determine the acceleration of the object. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 3.0 m/s

Distance (s) = 0.09 m

Acceleration (a) =?

v² = u² + 2as

3² = 0² + (2 × a × 0.09)

9 = 0 + 0.18a

9 = 0.18a

Divide both side by 0.18

a = 9 / 0.18

a = 50 m/s²

Finally, we shall determine the mass of the object. This can be obtained as follow:

Net Force (F) = 3 N

Acceleration (a) = 50 N

Mass (m) =?

F = ma

3 = m × 50

Divide both side by 50

m = 3 / 50

m = 0.06 Kg

Therefore, the mass of the object is 0.06 Kg

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3 years ago

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

V125BC [204]

D. 128.25 because force=mass x acceleration

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4 years ago

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During the spin cycle of a washing machine, the clothes stick to the outer wall of the barrel as it spins at a rate as high as 1

Darya [45]

To answer the two questions, we need to know two important equations involving centripetal movement:

v = ωr (ω represents angular velocity in radians)

a = $\frac{v^{2}}{r}$

Let's apply the first equation to question a:

v = ωr

v = ((1800*2π) / 60) * 0.26

Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.

Back to the equation:

v = ((1800*2π)/60) * 0.26

v = (1800*2(3.14)/60) * 0.26

v = (11304/60) * 0.26

v = 188.4 * 0.26

v = 48.984

v = 49 (m/s)

Now that we know the linear velocity, we can find the centripetal acceleration:

a = $\frac{v^{2}}{r}$

a = $\frac{49^{2}}{0.26}$