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3 years ago

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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

2 answers:

agasfer [191]3 years ago

6 0

Answer:

It would take 6 seconds before hitting the ground

Explanation:

ira [324]3 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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Answer:

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Explanation:

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Mathematically, Gravitational field is represented as,

g = F/m ..................... Equation 1.

m = F/g ..................... Equation 2.

Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.

From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

m = 785/9.8

m = 80.10 kg.

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Given: m = 80.10 kg, F = 298.

Substituting into equation 2

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3 years ago

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strojnjashka [21]

The answer would be d!

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To do that we will use the following formula:

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If we assume the initial velocity to be 0 we get:

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Now, we substitute the values:

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$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

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$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

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Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

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$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

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Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

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Now, we take the square root to both sides:

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Llana [10]

Answer:

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please see the file attached for more details.

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