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sergeinik [125]

2 years ago

14

During a circus performance, a 71.2-kg human cannonball is shot out of an 18.0-m-long cannon. If the human cannonball spends 0.7

85 s in the cannon, determine the average net force exerted on him in the barrel of the cannon. (Assume the acceleration is constant.)

1 answer:

Sophie [7]2 years ago

7 0

Answer:

F=ma = 71.2 * 58.42 = 4159.5 N

Explanation:

for FORCE we need a: acceleration (F=ma)

for acceleration we can apply the second equation of motion

s = ut + 0.5at^2

where

s = 18 m ( displacement)

u = 0 m/s (initial velocity)

t = 0.785 s (time)

18 = 0.5 * a * (0.785)^2

a = 58.42 m/s^2

now

F=ma = 71.2 * 58.42 = 4159.5 N

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At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

<u>Known Data</u>

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

<u>Time of the Thrown Rock</u>

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago

Name two kinds of longitudinal waves and explain how you know they are longitudinal.

dezoksy [38]

A transverse wave and a longitudinal wave.

Transverse:wave particles move at medium speed in perpendicular to the direction that the waves move

Longitudinal:wave particles move at medium speed in parallel to the direction that the wave moves

Hope this helps ^-^

8 0

2 years ago

A 0.49-kg cord is stretched between two supports, 7.8m apart. When one support is struck by a hammer, a transverse wave travels

katovenus [111]

To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,

$v = \sqrt{\frac{T}{\mu}}$

Here,

v = Velocity

$\mu$ = Linear density (Mass per unit length)

T = Tension

Rearranging to find the Period we have that

$T = v^2 \mu$

$T = v^2 (\frac{m}{L})$

As we know that speed is equivalent to displacement in a unit of time, we will have to

$T = (\frac{L}{t}) ^2(\frac{m}{L})$

$T = (\frac{7.8}{0.83})^2 (\frac{0.49}{7.8})$

$T = 5.54N$

Therefore the tension is 5.54N

8 0

2 years ago

7.) What is the relationship between kinetic and potential energy?

Arte-miy333 [17]

We know that potential energy is the energy that is stored within an object while kinetic energy is the energy that is in motion. The connection between the two is that potential energy transforms into kinetic energy

4 0

2 years ago