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Shtirlitz [24]
3 years ago
12

Would someone plz help a shister outttt

Physics
1 answer:
tiny-mole [99]3 years ago
8 0

Yaaaas, do you watch James Charles!?

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Which of these will be the correct relationship between work input and work output?
dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

8 0
2 years ago
The fulcrum of a first-class lever divides its 9.0 m arm into two sections—a 6.0 m arm and a 3.0 m arm. You place a rock weighin
nexus9112 [7]
For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3

Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N

Finally, Option A would be your correct answer.

Hope this helps!
6 0
2 years ago
What element x is most likely to react to form the compound xf5?
djyliett [7]
<span>antimony. It has +3,+5,-3 so yeah. the others carbon+2,+4,-4, chlorine +1,+5,+7,-1 but -1 is the most often so it isn't Cl, calcium +2.</span>
8 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
3 years ago
Objects 1 and 2 attract each other with a gravitational force of 45 units. If the mass of Object 1 is doubled, then the new grav
Novosadov [1.4K]

Explanation:

Fgravity = G*(mass1*mass2)/D²

so, if you double one of the masses, what does that do to our equation ?

Fgravitynew = G*(2*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 2* G*(mass1*mass2)/D² = 2* Fgravity

so, the correct answer will be 2×45 = 90 units.

4 0
2 years ago
Read 2 more answers
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