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Pachacha [2.7K]
4 years ago
12

Can sonebody help answer and explain this problem:the problem is in the picture.​

Physics
1 answer:
wel4 years ago
6 0

Explanation:

"Rank the magnitude of the torque the signs exert about the point at which the rod is attached to the side of the building."

Torque is the cross product of the radius vector and force vector.

τ = r × F

The magnitude of the torque is the force times the perpendicular distance.  In other words, it is the force times distance times the sine of the angle between them.

τ = F r sin θ

Let's assume the length of each rod is L meters.  In each case, the weight is a vertical force, so the perpendicular distance is the horizontal distance between the wall and the sign.

A) τ = (50 kg × 10 m/s²) (L sin 30° m)

τ = 250L Nm

B) τ = (100 kg × 10 m/s²) (L m)

τ = 1000L Nm

C) τ = (50 kg × 10 m/s²) (L sin 60° m)

τ ≈ 433L Nm

D) τ = (90 kg × 10 m/s²) (L m)

τ = 900L Nm

Ranked from greatest to least:

B > D > C > A

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Answer:

298800 m

Explanation:

v =  \frac{d}{t}

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d=distance

t=time

d = vt

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8.3m/s * 36000s

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What is saturns most distinctive feature?
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A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
Cerrena [4.2K]

Answer:

A) 112 m. B) 27.2 m C) 41.1 m/s i + 13.4 m/s j  D) 43.2 m/s

Explanation:

A) Once fired, no external forces act on the projectile in the x-direction, so it keeps moving to the right at constant speed, which is the projection on the x-axis of  the initial velocity vector:

v₀ₓ = v₀* cos 33º = 49 m/s* cos 33º = 41.1 m/s

In the y-direction, the component of the velocity can be found as the projection of v₀ on the y-axis, as follows:

v₀y = v₀* sin 33º = 49 m/s* sin 33º = 26.7 m/s

Both velocities are independent each other, as no one has a projection on the other.

In the vertical direction, the  projectile is in free fall all time, under the influence of gravity , which accelerates it downward.

So, at any time, in the vertical direction, the velocity can be calculated as follows:

vfy = v₀y -g*t (same equation as for an object thrown upwards)

When the object is at its maximum height, the velocity, in the vertical direction, will be momentarily zero, so we can find the time when this happens as follows:

vfy= 0 ⇒ v₀y = g*t ⇒ t = v₀y / g = 26.7 m/s / 9.8 m/s² = 2.72 s

As the time is the same for both movements, we can replace this value in the expression for the displacement x at constant speed, as follows:

x = v₀ₓ* t = 41.1 m/s* 2.72 s = 112 m

B) Like above, as the time is the same for both movements, we can find the time for the instant that the projectile hit the wall, as follows:

x = v₀ₓ* t ⇒ 55. 8 m = 41.1 m/s * t

⇒ t = 55. 8 m / 41.1 m/s = 1.36 s

We can replace this value of t in the equation for the vertical displacement, as follows:

Δy = v₀y*t -1/2*g*t² = (26.7m/s*1.36s) - 1/2*9.8m/s²*(1.36s)² = 27.2 m

C) The velocity of the projectile, at any time, has two components, one horizontal and one vertical.

As explained above, x-component is constant, equal to v₀x:

vx = v₀x i = 41.1 m/s i

For vy, we can apply acceleration definition, using the value of v₀y and t that we have just found:

vfy = voy - g*t = 26.7 m/s - 9.8m/s*1.36 sec = 13.4 m/s

vfy = 13.4 m/s j

v = 41.1 m/s i + 13.4 m/s j

D) Finally, in order to get the speed of the projectile when it hit the wall, we need just to find the magnitude of the velocity, as we get the magnitude of any vector given its vertical and horizontal components:

v = √(41.1 m/s)² +(13.4 m/s)² =43.2 m/s

5 0
3 years ago
PLEASE HELP ASAP URGENT!!!!!
coldgirl [10]

Answer:

B

i am not fully sure but i believe its the most accurate its either A or B

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3 years ago
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Answer:

Explanation:

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