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amm1812
3 years ago
8

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

lt a 2.0 kg robot "mouse" that runs up and down the rope. What minimum magnitude of the acceleration should the robot have for the rope to fail? Express your answer with the appropriate units.
Physics
1 answer:
raketka [301]3 years ago
8 0

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

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Can anyone help me with this? <br>the question : find the value of the resistors​
weeeeeb [17]

Answer:

Value of the resistors​ R = 2 ohm

Explanation:

Given:

Resistor R1 = 4 ohm

Resistor R2 = 4 ohm

Find:

Value of the resistors​ R

Computation:

We know that Resistor R1 and Resistor R2 connected in parallel series.

So,

Using common resistance in parallel series

⇒ 1/R = 1/R1 + 1/R2

⇒ 1/R = 1/4 + 1/4

⇒ 1/R = (1 + 1) / 4

⇒ 1/R = 2/4

⇒ R = 4/2

⇒ R = 2

Value of the resistors​ R = 2 ohm

3 0
3 years ago
How many meters is in 32 km
alina1380 [7]

Answer:

32000 m

Explanation:

1000m in 1 km, so 32000m in 32 km.

6 0
3 years ago
Read 2 more answers
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed
Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

\int EdS=\frac{Q_{in}}{\epsilon_o}   (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum =  8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

\int EdS=ES=E(4\pi r^2)   (2)

Qin is calculate by using the charge density:

Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho  (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}

Next, you use the results of (3), (2) and (1):

E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}

hence, the electric field is 1580594.95 N/C

7 0
3 years ago
If the Sun subtends a solid angle Ω on the sky, and the flux from the Sun just above the Earth’s atmosphere, integrated over all
Arada [10]

Answer:

A)Ω = 7.8 × 10^−5 steradians.

B) TE = 5800K

C) fλ(λ1) = (π ^2 ) /ΩBλ(T)

Explanation:

A) First of all, if we assume that the Sun emits isotropically at a luminosity (L⊙) , the flux at a given distance R from the sun would be f(d) = L⊙/ (4πd^2)

The ratio of flux at the solar photosphere to the flux at the Earth’s atmosphere would be: F⊙/{f(d⊙)} = (R⊙)^2 / (d⊙)^2

Now if we think of this relationship of the flux and the earth as a conical pattern, we'll deduce that the solid angle subtended by the sun at Earth’s surface to be;

Ω = π[(R⊙)^2 / (d⊙)^2]

Combining this with the ratio earlier gotten, well arrive at;

F⊙ = {f(d⊙ )π} /Ω

Now let's express The radius of the sun (R) in terms of its angular diameter (2α) and this gives;

R⊙ ≈ αd⊙

Now combining this with the equation for Ω earlier, we get;

Ω ≈ πα^2

So, = π((0.57/2π) /180)^2 = 7.8 × 10^−5 steradians.

B) from Stefan-Boltzmann Law,

F⊙ = σ(TE)^4

From the beginning, we know that;

F⊙ = {f(d⊙ )π} /Ω

And so replacing that in the stephan boltzmann law, we get ;

{f(d⊙ )π} /Ωσ = (TE)^4

So, (TE)^4 = {π (1.4 kWm^(−2))} / [(7.8 × 10^(−5 ) steradians x (5.66961 × 10^(−8))]

In stephan boltzmann law, σ = 5.66961 × 10^(−8)

And so, TE is approximately 5800K.

C) In order to relate fλ(λ1) with T, let's assume the sun’s surface to be an isotropically emitting blackbody, i.e its specific intensity is Iλ = Bλ(T). Hence, the flux at Sun’s surface for a given wavelength would be;

Fλ(λ1) = πBλ(T)

Now, if we combine this with the expression of F⊙ gotten earlier, well get the relation;

fλ(λ1) = (π ^2 ) /ΩBλ(T)

7 0
3 years ago
The ideal mechanical advantage of a machine reflects the increase or decrease in force there world be without friction, it is al
brilliants [131]

True: the ideal mechanical advantage of a machine is always greater than the actual mechanical advantage because all machines must overcome friction.

Explanation:

For a simple machine, it is possible to calculate two types of mechanical advantage:

1) The Ideal Mechanical Advantage (IMA) is given by

IMA=\frac{d_e}{d_r}

where

d_r is the resistance arm

d_e is the effort arm

The IMA gives the mechanical advantage of the machine if there are no friction forces acting on it, and if all the work in input is converted into work in output with no loss of energy

2) The Actual Mechanical Advantage (AMA) is given by

AMA=\frac{L}{E}

where

L is the load (the force in output)

E is the effort (the force in input)

The AMA gives the real mechanical advantage of the machine. For an ideal machine,

AMA=IMA

Because there is no loss of energy due to friction.

For a real machine instead,

AMA

because part of the input energy is converted into thermal energy and other forms of energy due to the presence of friction, so it is "wasted" energy.

Learn more about levers and machines:

brainly.com/question/5352966

#LearnwithBrainly

8 0
4 years ago
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