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Scrat [10]
3 years ago
11

for the meter stick shown in figure 10-4, the force F1 10.0N acts at 10.0cm. what is the magnitude of the torque due to f1 about

an axis through point A perpendicular to the page?
Physics
1 answer:
Solnce55 [7]3 years ago
5 0

The magnitude of the torque due to f1 about an axis through point A can be calsulated using

  t = Fd

where t is the torque

f is the force applied

and d is the distance

t = (10 n) x ( 0.10 m)

t = 1 Nm

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In your own words, explain the effects of changes in Earth's magnetic field over time.
viva [34]
It's an interesting fact that scientists don't fully understand how it works. But it seems to be to do with molten metal circulating in the core. Given that it's just liquid metal sloshing around, it seems understandable that it won't always circulate perfectly - imagine the cloud bands in Jupiter's atmosphere - they are reasonably stable but change from time to time. When the liquid changes its speed or direction, however slowly it does so, the resulting magnetic field will move or switch direction.

<span> As scientists try to build better mathematical models of how the core works, they should be able to learn more about the magnetic field it produces. Hope this helps</span>
3 0
3 years ago
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__________ helps us adapt to our environment. It also generally __________ with age. A. Plasticity . . . decreases B. Developmen
Nana76 [90]
Not sure this is a physics question (probably biology).
Anyway, the correct answer is A):
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8 0
3 years ago
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Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to
marta [7]

Answer:

\frac{ 112.5}{15+m_{A}}=v_{f}

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed v_{iA} of 0 m/s and a mass m_{A} and the IMAX camera with an initial speed v_{ic} of 7.5 m/s and a mass m_{c} of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

3 0
3 years ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 mpa m (36.4 ksi in.
GarryVolchara [31]
We will first determine using the given if an aircraft component will fracture with a given stress level (260 MPa), maximum internal crack length (6.0 mm) and fracture toughness (40 MPa m ), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. First, it is necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and 2 a = 4.0 mm). Therefore,

 Y = K(Ic)/ sqrt(π a) = 40 MPa( m ) / (300 MPa) sqrt(( π ) ((4 × 10-3 m)/2)) = 1.68 

We will now solve for the product Y σ π a for the other set of conditions, so as to ascertain whether or not this value is greater than the K(Ic) for the alloy. Thus, 

Y sqrt(π a) = (1.68)(260 MPa) sqrt (( π )[(6 × 10^-3 m)/ 2])
                  = 42.4 MPa sqrt (m) (39 ksi in. ) 

Therefore, fracture will occur since this value ( 42.4 MPa sqrt(m)) is greater than the K(Ic) of the material, 40 MPa sqrt(m).
8 0
4 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
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