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romanna [79]
2 years ago
5

4. Ano ang matinding suliranin ang kakaharapin kapag patuloy ang pagkasira ng

Physics
2 answers:
Cerrena [4.2K]2 years ago
7 0
It’s D :)))))))))))))))))
jasenka [17]2 years ago
5 0

Answer:

I think its D

Explanation:

Kasi Pinoy ako

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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

5 0
2 years ago
Janelle is exploring the relationship between the brightness of a light bulb and the current that powers it. When applying these
Vlada [557]

Answer:B

Explanation:

Y is the brightest bulb

7 0
3 years ago
Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
3 years ago
The ice cap at the North Pole may be 1000 m thick, with a density of 920 kg/m3. Find the pressure at the bottom and the correspo
bija089 [108]
<span>Pice=920kg/m^3 deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa P=Po + deltaP=101.325 + 9022 = 9123kPa</span>
7 0
2 years ago
19) A child on a sled starts from rest at the top of a 15.0° slope. If the trip to the bottom takes 15.2 s,
sveticcg [70]

Answer: 288.8 m

Explanation:

We have the following data:

t=15.2 s is the time it takes to the child to reach the bottom of the slope

V_{o}=0 is the initial velocity (the child started from rest)

\theta=15\° is the angle of the slope

d is the length of the slope

Now, the Force exerted on the sled along the ramp is:

F=ma (1)

Where m is the mass of the sled and a its acceleration

In addition, if we draw a free body diagram of this sled, the force along the ramp will be:

F=mg sin \theta (2)

Where g=9.8 m/s^{2} is the acceleration due gravity

Then:

ma=mg sin \theta (3)

Finding a:

a=g sin \theta (4)

a=9.8 m/s^{2} sin(15\°) (5)

a=2.5 m/s^{2} (6)

Now, we will use the following kinematic equations to find d:

V=V_{o}+at (7)

V^{2}=V_{o}^{2}+2ad (8)

Where V is the final velocity

Finding V from (7):

V=at=(2.5 m/s^{2})(15.2 s) (9)

V=38 m/s (10)

Substituting (10) in (8):

(38 m/s)^{2}=2(2.5 m/s^{2})d (11)

Finding d:

d=288.8 m

6 0
3 years ago
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