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Arte-miy333 [17]
2 years ago
10

4. We have 4 identical strain gauges of the same initial resistance (R) and the same gauge factor (GF). They will be used as R1,

R2, R3, and R4 in a Wheatstone bridge. We want to measure the tensile strain of a rod subjected to a tensile force. What is the maximum bridge constant (kappa) that we can reach
Physics
1 answer:
Luba_88 [7]2 years ago
3 0
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Two friends of different masses are on the playground. They are playing on the seesaw and are able to balance it even though the
Westkost [7]

Answer:

They are able to balance torques due to gravity.

F_1 L_1 = F_2L_2

Explanation:

When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal

This condition will be torque equilibrium position where the see saw will not rotate

Here we can say

F_1 L_1 = F_2L_2

here we know that force is due to weight of two friends

and their positions are different with respect to the lever about which see saw is rotating

since both friends are of different weight so they will balance themselves are different positions as per above equation

5 0
2 years ago
Ever tried to stop a 150-pound (68 kg) cannonball fired towards you at 30 mph (48 km/hr.)? No, probably not. But you may have tr
alex41 [277]
The two situations are similar because in both you are trying to minimize the damage and make the best out of a bad situation
8 0
2 years ago
explain why the impact of one heavy stone would produce waves with higher amplitude than the impact of the light stone would
DerKrebs [107]
The heavy stone would produce waves with a higher amplitude, rather than the smaller stone, because since the stone is heaver its going to have a grater impact and displace more water to create a bigger wave.
5 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
2 years ago
Grain is pored into a silo to be stored for later use. Due to the friction between pieces of grain as they rub against eachother
MariettaO [177]

Answer:

3.1×10⁻¹¹ N

Explanation:

Use Coulomb's law:

F = k q₁ q₂ / r²

F = (9×10⁹) (6.0×10⁻¹⁰) (2.3×10⁻¹⁵) / (0.02 m)²

F = 3.1×10⁻¹¹

6 0
2 years ago
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