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3 years ago

3. A model rocket is launched straight upward at 58.8 m/s.

Physics

1 answer:

kolezko [41]3 years ago

8 0

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t = $\frac{58.8}{9.8}$

= 6 seconds

b). Total time of flight = 2(Time taken to go up)

= 2×6

= 12 sec

c). Formula to get the peak height is,

$h=ut-\frac{1}{2}gt^2$

h = (58.8)6 - $\frac{1}{2}(9.8)(6)^2$

= 352.8 - 176.4

= 176.4 meters

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If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is

viktelen [127]

To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
where
$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
where
$d$ is the distance
$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
$t=0.98$ seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
$d=V_{i}t+ \frac{1}{2} gt^2$
$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
$d=15.22$ ft

Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
$16t^2+31.21t-175.22=0$
$t=2.47$ or $t=-4.43$
Since time cannot be negative, $t=2.47$ is the time it takes the object to fall to the ground.

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
$V_{f}=V_{i}+gt$
$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

5 0

2 years ago

A 70 kg human sprinter can accelerate from rest to 10 m/s in 3.0 s. During the same time interval, a 30 kg greyhound can go from

ladessa [460]

Answer:

$P_1 = 1166.7 Watt$

$P_2 = 2000 Watt$

Explanation:

Average power for the human sprinter is given as

$Power = \frac{\Delta E}{\Delta t}$

so we have

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(70)(10^2) - 0}{3}$

$P_1 = 1166.7 Watt$

Average power for greyhound is given as

$P = \frac{\frac{1}{2}mv^2 - 0}{\Delta t}$

$P = \frac{\frac{1}{2}(30)(20^2) - 0}{3}$

$P_2 = 2000 Watt$

3 0

3 years ago

A scuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. At what angl

LUCKY_DIMON [66]

Answer:

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

180 = θ2 + 90 + θ_p

θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection incident angle equals reflected angle,

ni sin θ_p = ns cos θ_p

n₂ / n₁ = sin θ_p / cos θ_p

n₂ / n₁ = tan θ_p

θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

θ_p = tan⁻¹ (1.33 / 1)

θ_p = 53.0º

n₂ = 1.40

θ_p = tan⁻¹ (1.40 / 1)

Tep = 54.5º

4 0

3 years ago

The weight of an apple at 4r from earth is of that on earth

iris [78.8K]

W = m.g = weight
g = Gme/Re**2 where G= universal gravitational constant , Re= radius of the earth
me= mass of the earth
therefore it weighs 16 times less

4 0

2 years ago

Read 2 more answers

A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned

Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, $\omega _i$ = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, $\omega_f$ = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

$\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\ -\ \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\ -\ (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad$

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0

2 years ago