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2 years ago

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Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

e show work!

1 answer:

Westkost [7]2 years ago

8 0

Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi

Orlov [11]

The number converted is $0.0467 \frac{(kg)(s)}{m^3}$

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

$1 kg = 1000 g = 10^6 mg$

$1 min = 60 s$

$1 m^3 = 1000 L$

The original unit that we have is

$\frac{mg\cdot min}{L}$

Therefore, it can be rewritten as:

$=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot 60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}$

Therefore, since the initial number was 0.779, the final value is

$0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}$

#LearnwithBrainly

5 0

3 years ago

Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l

wlad13 [49]

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2

Since the Ms are the same, then

2mx^2 / L^2 = m(L - x)^2 / L^2

2x^2 = (L - x)^2

xsqrt2 = L - x

x(1 + sqrt2) = L

x = L / (sqrt2 + 1) From here, we rationalize.

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)

x = L(sqrt2 - 1) / (2 - 1)

x = L(sqrt2 - 1) </span>

= 0.414L

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

$I_1\omega_1=I_2\omega_2$

Here, $I_2=I_1+mr^2$

$I_1\omega_1=(I_1+mr^2)\omega_2$

$900\times 0.95=(900+70\times (2.9)^2)\omega_2$

Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

$k_i=\dfrac{1}{2}I_1\omega_1^2$

$k_i=\dfrac{1}{2}\times 900\times (0.95)^2$

$k_i=406.12\ rad/s$

The final rotational kinetic energy is given by :

$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

5 0

3 years ago

a diver of mass 101 kg jumps upward off a diving board into water. Diving board is 6m above water. Diver has a speed of 1.2m/s.

8090 [49]

When the diver reaches maximum height, the upward velocity will be zero.

We shall use the formula
v^2 = u^2 - 2gh
where
v = 0 (velocity at maximum height)
u = 1.2 m/s, intial upward velocity
g = -9.8 m/s^2, gravitational acceleration (downward)
h = maximum height attained above the diving board.

Therefore
0 = 1.2^2 - 2*9.8*h
h = 1.2^2/(2*9.8) = 0.0735 m

Answer: 0.074 m (nearest thousandth)

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3 years ago

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o-na [289]

I bet she does just give her tule work on yourself

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3 years ago

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