the answer is 40.5 because you have to multiply the density and volume of the object to get the mass.
Answer:
F. The length of day would be longer
Explanation:
- Earths rotates on its axis from the eastward to the westward direction, earth takes about 24 hours to complete its rotation. If the rate of earth's rotation on its axis slows down then the days will be longer than usual.
- Thought the length of year would remain the same. Earth rotation is also impacted by the sun and moon's gravitational pull. The disappearance of the centrifugal force form the surface would cause climate changes and seasonal extremes.
Answer:
a) τ₁ = 660 N m, b) τ’= 686 N m, c) F = 623.6 N
Explanation:
a) For this exercise let's use the concepts of torque and rotational balance.
For this we set a reference system at the base and assuming that the counterclockwise rotations are positive
where the force F = 600 N, the distance to the axis is x = 1.1 m, the mass of the system m = 70g and the weight is placed at the point of the center of gravity x_{cm} = -1.0 m
The torque at the front is
τ₁ = F x
τ₁ = 600 1.1
τ₁ = 660 N m
b) let's write the rotational equilibrium condition
∑ τ = 0
τ'- W x_{cm} = 0
τ ’= mg x_{cm}
τ’= 70 9.8 1.0
τ’= 686 N m
c) the greatest force Matt can apply
τ’= F x
F = τ’/ x
F = 686 / 1.1
F = 623.6 N
Answer:
Explanation:
Let the amplitude of individual wave be I and resultant amplitude be 1.703 I . Let the phase difference be Ф in terms of degree
From the formula of resultant vector
(1.703I)² = I² + I² + 2 I² cosФ
2.9 I² = 2I² + 2 I² cosФ
.9I² = 2 I² cosФ
cosФ = .9 / 2
= .45
Ф = 63.25 .
Answer:
<em>The range is 35.35 m</em>
Explanation:
<u>Projectile Motion</u>
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
Being vo the initial speed of the object, θ the initial launch angle, and
the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:
![\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20d%3D%7B%5Cfrac%20%20%7Bv_o%5E%7B2%7D%5Csin%282%5Ctheta%20%29%7D%7Bg%7D%7D)
The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:
![\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20d%3D%7B%5Cfrac%20%20%7B20%5E%7B2%7D%5Csin%282%5Ccdot%2030%5E%5Ccirc%20%29%7D%7B9.8%7D%7D)
![\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20d%3D%7B%5Cfrac%20%20%7B400%5Csin%2860%5E%5Ccirc%20%29%7D%7B9.8%7D%7D)
![d=35.35\ m](https://tex.z-dn.net/?f=d%3D35.35%5C%20m)
The range is 35.35 m