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2 years ago

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is

Physics

1 answer:

Natali5045456 [20]2 years ago

4 0

Answer:

Explanation:

54.34+45.66=100.00 (you have to use the .00 because when you add to numbers you keep the number of decimals)

so you get 100.00

all numbers that are not 0 are sig figs so 1 is a sig fig

If a number ends with a 0 after a decimal place that 0 is a sig fig

all numbers between two sig figs are sig figs so that would make all of the numbers sig figs

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A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte

Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0

2 years ago

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

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

The vertical distance is $d = \frac{2}{k} *[mg + f]$

Explanation:

From the question we are told that

The mass of the cylinder is m

The kinetic frictional force is f

Generally from the work energy theorem

$E = P + W_f$

Here E the the energy of the spring which is increasing and this is mathematically represented as

$E = \frac{1}{2} * k * d^2$

Here k is the spring constant

P is the potential energy of the cylinder which is mathematically represented as

$P = mgd$

And

$W_f$ is the workdone by friction which is mathematically represented as

$W_f = f * d$

$\frac{1}{2} * k * d^2 = mgd + f * d$

=> $\frac{1}{2} * k * d^2 = d[mg + f ]$

=> $\frac{1}{2} * k * d = [mg + f ]$

=> $d = \frac{2}{k} *[mg + f]$

5 0

2 years ago

The Orion nebula is one of the brightest diffuse nebulae in the sky (look for it in the winter, just below the three bright star

Oxana [17]

Answer:

T=183.21K

Explanation:

We have to take into account that the system is a ideal gas. Hence, we have the expression

$PV=nRT$

where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the ideal gas constant.

Thus, it is necessary to calculate n and V

V is the volume of a sphere

$V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (5.9*10^{15}m)^3=8.602*10^{47}m^3$

V=8.86*10^{50}L

and for n

$n=\frac{(4000M_s)/(2*mH)}{6.022*10^{23}mol^{-1}}=3.95*10^{36}mol$

Hence, we have (1 Pa = 9.85*10^{-9}atm)

$T=\frac{PV}{nR}=\frac{(6.8*10^{-9}*9.85*10^{-6}atm)(8.86*10^{50}L)}{(0.0820\frac{atm*L}{mol*K})(3.95*10^{36}mol)}\\\\T=183.21K$

hope this helps!!

8 0

2 years ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is​

In the sum of 54.34 and 45.66, the number of significant figure for theanswer is