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3 years ago

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a model rocket flies horizontally off the edge of a cliff at a velocity of 40.0m/s. if the canyon below is 110.0m deep, how far

from the edge of the cliff does the model rocket land?

1 answer:

Over [174]3 years ago

5 0

Answer:

189.6 m

Explanation:

First of all, we study the vertical motion of the rocket in order to find the time it takes for it to land. The suvat equation for the vertical motion is

$s=ut+\frac{1}{2}gt^2$

where, taking downward as positive direction:

s = 110.0 m is the vertical displacement

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(110)}{9.8}}=4.74 s$

Now we can just analyze the horizontal motion, which is a uniform motion with constant velocity, which is equal to

$v_x = 40.0 m/s$

So the distance travelled horizontall after a time t is

$d=v_x t$

So, when the rocket lands (t = 4.74 s), the horizontal distance travelled is

$d=(40.0)(4.74)=189.6 m$

Therefore, the rocket lands 189.6 m far from the edge of the cliff.

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A spherical balloon has a radius of 7.40 m and is filled with helium. Part A How large a cargo can it lift, assuming that the sk

malfutka [58]

Answer:

The mass of the cargo is $M = 188.43 \ kg$

Explanation:

From the question we are told that

The radius of the spherical balloon is $r = 7.40 \ m$

The mass of the balloon is $m = 990\ kg$

The volume of the spherical balloon is mathematically represented as

$V = \frac{4}{3} * \pi r^3$

substituting values

$V = \frac{4}{3} * 3.142 *(7.40)^3$

$V = 1697.6 \ m^3$

The total mass the balloon can lift is mathematically represented as

$m = V (\rho_h - \rho_a)$

where $\rho_h$ is the density of helium with a value of

$\rho_h = 0.179 \ kg /m^3$

and $\rho_a$ is the density of air with a value of

$\rho_ a = 1.29 \ kg / m^3$

substituting values

$m = 1697.6 ( 1.29 - 0.179)$

$m = 1886.0 \ kg$

Now the mass of the cargo is mathematically evaluated as

$M = 1886.0 - 1697.6$

$M = 188.43 \ kg$

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3 years ago

ASAP ANSWER PLEASE WILL MARK BRAINLIEST

lisov135 [29]

Yes

Explanation:

It is a reasonable result obtained.

Error = true value - measured value

true value = 24.5

measured value = 24.2

Error = 24.5 - 24.2 = 0.3g

The error reported in the reading is 0.3g

The reason why we had a disparity in the figures obtained from this measurement is primarily due to some erroneous scale.

The mixture at the end of the day is a solution.

We are expected to have the same mass but due to experimental or some form of random error introduced, we noticed a difference.

The value obtained is quite logical as we only had a deviation of 0.3g.

learn more:

Error brainly.com/question/2764830

#learnwithBrainly

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3 years ago

Most reactive non-metals in order?

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Carbon
</span> Nitrogen
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3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span> $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span> $E=K_{max} = \frac{1}{2}mv_{max}^2$
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span> $E=U_{max}= \frac{1}{2}k A^2$
<span>
Since the mechanical energy must be conserved, we can write
</span> $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
<span>from which we find the maximum speed
</span> $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span> $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
<span>And since
</span> $E=U+K$
<span>we find the kinetic energy when the glider is at this position:
</span> $K=E-U=0.36 J - 0.05 J = 0.31 J$
<span>And then we can find the corresponding velocity:
</span> $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span> $a(t)=\omega^2 x(t)$
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span> $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

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3 years ago

A biker pedals hard for 3 seconds. What is his initial velocity if he accelerated by 4m/s2 until he's going 20m/s. (Which equati

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Answer:

answer is option 4

Explanation:

you have to use option 4 because u need to find out initial velocity (Vi)

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3 years ago