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2 years ago

A screw-jack used to lift a bus is a

Physics

1 answer:

Gekata [30.6K]2 years ago

3 0

Second order lever broski

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In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

2 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position is given by

$\Delta h=L(1-\cos \theta )$

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

$mgL(1-\cos \theta )=\frac{mv^2}{2}$

$v=\sqrt{2gL(1-\cos \theta )}$

$v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}$

$v=\sqrt{4.021}$

$v=2 m/s$

5 0

2 years ago

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2

kkurt [141]

Answer:

$4.3\cdot 10^{-12} J$

Explanation:

The fusion reaction in this problem is

$4^1_1H \rightarrow ^4_2He +2e^+$

The total energy released in the fusion reaction is given by

$\Delta E = c^2 \Delta m$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

$m(^1_1H)=1.007825u$ is the mass of one nucleus of hydrogen

$m(^4_2 He)=4.002603u$ is the mass of one nucleus of helium

So the mass defect is:

$\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u$

The conversion factor between atomic mass units and kilograms is

$1u=1.66054\cdot 10^{-27}kg$

So the mass defect is

$\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg$

And so, the energy released is:

$\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J$

6 0

2 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

sergejj [24]

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

3 0

2 years ago