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3 years ago

15

True or false the mechanical advantage to simple machines is that they allow a decreased input force to create a larger output f

orce.

1 answer:

Fittoniya [83]3 years ago

4 0

I think its true im not sure

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Convert 2536 mm/min to m/s. Use dimensional analysis.

Deffense [45]

Answer:

<u>As</u><u> </u><u>we</u><u> </u><u>kno</u><u>w</u><u> </u><u>that</u><u>,</u><u> </u>

1 mm/min = 1.66667E-5 m/s
1 m/s = 60000 mm/min

<u>Now</u><u>,</u><u> </u><u>come</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>question</u><u> </u><u>-</u><u> </u>

$\\ \implies \sf \: 2536 \times 1.66667E-5 m/s \\ \\ \\ \implies \sf \blue{0.0422666667 \:m/s } \\$

Result : 2536 mm/min = 0.0422666667 m/s.

6 0

4 years ago

You and your friends are pushing a car with a total of 1080 N of force. You are able to push the car for a distance of 218 m. Ca

harkovskaia [24]

Answer:

How much power is required to pull a sled if you use 60J of work in 5 seconds? ... Misfortune occurs and Renatta and her friends find themselves getting a workout. They apply a cumulative force of 1080 N to push their car 218 m to the ... Calculate the amount of work done when moving a 567N crate a distance of 20 meters.

Explanation:

Misfortune occurs and Renatta and her friends find themselves getting a ... They apply a cumulative force of 1080 N to push the car 218 m to the nearest fuel ... Write down what they give you. ... Determine the work done by Lamar in deadlifting 300 kg to a height of 0.90 m ... Work = Force x Distance = Joules (force = weight).

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3 years ago

Cross product and its properties

bearhunter [10]

The length of the cross product of two vectors

The scalar triple product of the vectors a, b, and c

The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product.

<u>Explanation</u>:

The length of the cross product of two vectors is | a $\times$ b | = |a| |b| sin θ

The length of the cross product of two vectors is equal to the area of the parallelogram determined by the two vectors (see figure below).

Anticommutativity:

| a $\times$ b | = - | b $\times$ a |

Multiplication by scalars:

(ca) $\times$ b = c (a $\times$ b) = a $\times$ (cb)

Distributivity:

a $\times$ (b + c) = (a $\times$ b) + (a $\times$ c)

The scalar triple product of the vectors a, b, and c:

a . (b $\times$ c) = (a $\times$ b) . c

The magnitude of the scalar triple product is the volume of the parallelepiped of the vectors a, b, and c.

The vector triple product of the vectors a, b, and c is given as

a $\times$ (b $\times$ c) = (a.c) b - (a.b) $\times$ c

4 0

4 years ago

A 0.42 kg mass is attached to a light spring with a force constant of 34.9 N/m and set into oscillation on a horizontal friction

Whitepunk [10]

(a) 0.456 m/s

The maximum speed of the oscillating mass can be found by using the conservation of energy. In fact:

- At the point of maximum displacement, the mechanical energy of the system is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$ (1)

where

k = 34.9 N/m is the spring constant

A = 5.0 cm = 0.05 m is the amplitude of the oscillation

- At the point of equilibrium, the displacement is zero, so all the mechanical energy of the system is just kinetic energy:

$E=K=\frac{1}{2}mv_{max}^2$ (2)

where

m = 0.42 kg is the mass

vmax is the maximum speed, which is maximum when the mass passes the equilibrium position

Since the mechanical energy is conserved, we can write (1) = (2):

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{(34.9 N/m)(0.05 m)^2}{0.42 kg}}=0.456 m/s$

(b) 0.437 m/s

When the spring is compressed by x = 1.5 cm = 0.015 m, the equation for the conservation of energy becomes:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (3)

where the total mechanical energy can be calculated at the point where the displacement is maximum (x = A = 0.05 m):

$E=\frac{1}{2}kA^2=\frac{1}{2}(34.9 N/m)(0.05 m)^2=0.044 J$

So, solving (3) for v, we find the speed when x=1.5 cm:

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(c) 0.437 m/s

This part of the problem is exactly identical to part b), since the displacement of the mass is still

x = 1.5 cm = 0.015 m

So, the speed when this is the displacement is

$v=\sqrt{\frac{2E-kx^2}{m}}=\sqrt{\frac{2(0.044 J)-(34.9 N/m)(0.015 m)^2}{0.42 kg}}=0.437 m/s$

(d) 4.4 cm

In this case, we have that the speed of the mass is 1/2 of the maximum value, so:

$v=\frac{v_{max}}{2}=\frac{0.456 m/s}{2}=0.228 m/s$

And by using the conservation of energy again, we can find the corresponding value of the displacement x:

$E=\frac{1}{2}kx^2+\frac{1}{2}mv^2\\x=\sqrt{\frac{2E-mv^2}{k}}=\sqrt{\frac{2(0.044 J)-(0.42 kg)(0.228 m/s)^2}{34.9 N/m}}=0.044 m=4.4 cm$

4 0

3 years ago

Jared would describe a square as having four equal sides and four right angles. This is Jared’s __________ of a square. A. model

zheka24 [161]

Answer:

D

Explanation:

jhusskgtyddutsuudtiff

4 0

3 years ago