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stich3 [128]
3 years ago
6

sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

from a point 6.80 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground

Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

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A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra
blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

7 0
3 years ago
On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the
IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = \frac{1}{2}\varepsilon _{0}E^{2}

          =\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119

          = 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = \frac{B^{2}}{2\mu _{0}}

=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0
3 years ago
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si
solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0
3 years ago
What happens when you "crack/pop your knuckles?
Art [367]

It could result in it not being good for your joints, as well as in the long run but shouldn't cause problems when your a child. I hope this helps your question!

6 0
3 years ago
In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex
sleet_krkn [62]
The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r 

g - v²/r = 0 

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s
8 0
3 years ago
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