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3 years ago

15

A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second. What is the final velocity of the b

all after 6.0 seconds?

1 answer:

a_sh-v [17]3 years ago

5 0

19.2 meters/second^2 would be the correct answer.

Try multiplying 3.2 meters/second^2 by 6 and you will receive the answer provided above. If you have any further questions, let me know!

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A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

b)

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

<u>height fallen freely by Austin:</u>

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

<u>Velocity just before opening the parachute:</u>

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

<u>Time taken by the helicopter to fall:</u>

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

<u>Now the height fallen in the remaining time using parachute:</u>

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

2 years ago

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

3 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

What is the new acceleration (in m/s/s)?

Delicious77 [7]

The new acceleration is $108 m/s^2$

Explanation:

We can answer this problem by applying Newton's second law, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

The equation can be rewritten as

$a=\frac{F}{m}$

In this problem, the initial acceleration is

$a=18.0 m/s^2$

Later:

- The net force is tripled: $F'=3F$

- The mass is halved: $m'=\frac{m}{2}$

Therefore, the new acceleration is:

$a'=\frac{F'}{m'}=\frac{3F}{m/2}=6\frac{F}{m}=6a$

which means that the new acceleration is 6 times the original acceleration, therefore

$a'=6(18)=108 m/s^2$

Learn more about acceleration:

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brainly.com/question/11181826

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#LearnwithBrainly

8 0

3 years ago

Which of the following is an example of speed?

Arturiano [62]

The cat has a speed. The truck and the bicycle both have velocity. We can't be sure of what the plane has.

6 0

3 years ago