Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

Answer:

T=12544 N*m

Explanation:

Given

L=4.0m

ms=500kg

mw=70kg

Torque is the force in a distance the relation is proportional so the torque of weight first is:

Ts = Fs*d

Ts = ms*g*L

Ts = 500kg*9.8m/s^2*2m

Ts = 9800 N*m

now torque of the worker

Tw = Fw*d

Tw = 70kg*9.8m/s^2*4m

Tw = 2744 N*m

Torque net is

Tnet = Tw+Ts

Tnet= 2744 + 9800 =12544 N*m

Final answer:

The total torque about the bolt due to the worker and the weight of the beam is 12544 Nm. This is found by adding the torque due to the beam and the worker which can be calculated using their weights and their distance from the pivot point (bolt).

Explanation:

The key to solving this question is understanding torque, which in physics represents the rotational effect of a force. Torque is calculated using the formula τ = r x F, where τ is the torque, r is the distance from the pivot point, and F is the force applied.

In this case, there are two forces to consider: the weight of the beam and the weight of the worker. Both of these can be calculated using the formula for weight (F = m*g), where m is mass and g is gravitational acceleration, which is approximately 9.8 m/s^2 on Earth. The weight of the beam is therefore 500 kg * 9.8 m/s^2 = 4900 N, and the weight of the worker is 70 kg * 9.8 m/s^2 = 686 N.

The distance from the pivot (bolt) for the beam's weight is considered to be the midpoint of the beam, so it is 4.0 m / 2 = 2.0 m. For the worker, r equals the full length of the beam, which is 4.0 m. The total torque can be calculated by adding the torque due to the beam and the worker. Therefore, the total torque τ = (2.0 m * 4900 N) + (4.0 m * 686 N) = 9800 Nm + 2744 Nm = 12544 Nm.

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Data given:

V=45m/s

t=11s

Δx=?

Formula needed:

V=Δx/t

Solution:

Δx=v×t

Δx=45m/s×11s

Δx=495m

According to my solution B) is the most accurate

Answer:

3.27 turns

Explanation:

To find how many turns (θ) will the stone make before coming to rest we will use the following equation:

Where:

: is the final angular velocity = 0

: is the initial angular velocity = 71.150 rpm

α: is the angular acceleration

First, we need to calculate the angular acceleration (α). To do that, we can use the following equation:

Where:

I: is the moment of inertia for the disk

τ: is the torque  

The moment of inertia is:

Where:

m: is the mass of the disk = 105.00 kg

r: is the radius of the disk = 0.297 m

Now, the torque is equal to:

Where:

F: is the applied force = 46.650 N      

μ: is the kinetic coefficient of friction = 0.451

The minus sign is because the friction force is acting opposite to motion of grindstone.    

Having the moment of inertia and the torque, we can find the angular acceleration:

Finally, we can find the number of turns that the stone will make before coming to rest:

I hope it helps you!

The rotation of Uranus, like that of Venus, is retrograde and its axis of rotation is inclined almost ninety degrees above the plane of its orbit. During its orbital period of 84 years one of the poles is permanently illuminated by the Sun while the other remains in the shade. Exactly its rotation period is equivalent to 17 hours and 14 Earth minutes and its translation period is equivalent to 84 years, 7 days and 9 Earth hours.

Only a narrow band around the equator experiences a rapid cycle of day and night, but with the Sun very low on the horizon as in the polar regions of the Earth. On the other side of the orbit of Uranus, the orientation of the poles in the direction of the Sun is inverse. Each pole receives about 42 years of uninterrupted sunlight, followed by 42 years of darkness. Therefore an observer at latitude of 45 degrees in Uranus will probably experience a long winter night that is equivalent to one third of the year uranium.

Answer:

v(t) = 21.3t

v(t) = 5.3t

Explanation:

When no sliding friction and no air resistance occurs:

where;

Taking m = 3 ; the differential equation is:

By Integration;

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

Taking m =3 ; the differential equation is;

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

The differential equation is :

=

=

By integration

Since; V(0) = 0 ; Then C = -48

A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².  

v² - u² = 2 a ∆x, where u and v are initial and final velocities, respectively; a is acceleration.

and ∆x is the distance traveled (because the skater moves in only one direction).

Thus, (5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s².

Thus, A speed skater moving across frictionless ice at 8.8 m/s hits a 6.0 m -wide patch of rough ice. Her acceleration on the rough ice is -3.65 m/s².  

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Recall that

v² - u² = 2 a ∆x

where u and v are initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled (because the skater moves in only one direction).

So we have

(5.8 m/s)² - (8.8 m/s)² = 2 a (6.0 m)

a = ((5.8 m/s)² - (8.8 m/s)²) / (12 m)

a = -3.65 m/s²