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3 years ago

Which of the following is an example of sound energy

1 answer:

4 0

Sound energy is produced when an object vibrates so an example would be a telephone ringing or someone playing a bass guitar

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A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to

NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago

A message is sent from the Galileo spacecraft orbiting Jupiter to earth at a distance of 928,000,000km. If it took the signal 51

expeople1 [14]

<span>The answer would approximately be 299,741.60</span>

6 0

3 years ago

Read 2 more answers

Suppose a car of mass m is moving at a constant speed v of

SIZIF [17.4K]

Answer:

The angle of banked curve that makes the reliance on friction unnecessary is

$\arcsin(v^2/(gR))$

Explanation:

In order the car to stay on the curve without friction, the net force in the direction of radius should be equal or smaller than the centripetal force. Otherwise the car could slide off the curve.

The only force in the direction of radius is the sine component of the weight of the car

$w_r = mg\sin(\theta)$