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chubhunter [2.5K]
4 years ago
9

What type of stress is shown in the following diagram?

Physics
2 answers:
leva [86]4 years ago
7 0
Here stress is parallel to the surface of the body. So it's a Shear stress.
Makovka662 [10]4 years ago
4 0
I did the test it is b
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3 years ago
A railroad car of mass 2.50 3 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railro
ipn [44]

Answer:2.5 m/s

37.5KJ

Explanation:

Let u_1, u_2 , v_f be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.

m=2.50\times 10^{4}

Conserving momentum

mu_1+3mu_2=4mv_f

4m+6m=4mv_f

v_f=2.5 m/s

Therefore Final velocity of system is 2.5m/s

(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy

Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J

Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J

Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ

4 0
3 years ago
Consider a particle with unit charge q, and mass m, in a constant magnetic field B directed along the positive z–axis. The parti
max2010maxim [7]

Answer:

it must be helical motion in which the charge particle will move uniformly along z axis and simultaneously it will move in circular path in xy plane.

Explanation:

Magnetic field is along z axis while velocity is in x-z plane

so we will have

F = q(\vec v \times \vec B)

so here we can say

F = q(u\hat i + w\hat k) \times (B \hat k)

so we will have

F = quB(-\hat j)

so here the net force on the charge is perpendicular to its x directional velocity along - Y direction

So due to this component of motion it will move along a circle while other component of the motion will remain uniform always

So here it is combination of two parts

1) Uniform circular motion

2) Uniform motion

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3 years ago
True or false compounds only have one type of element
klasskru [66]

False, Compounds can contain more than 2 elements.

8 0
3 years ago
If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?
Novay_Z [31]
Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>
8 0
3 years ago
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