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3 years ago

If a 10 meter ramp helps you move a 500 kg object up 1 meter. What was the mechanical advantage of the ramp?

1 answer:

6 0

The mechanical advantage of an inclined plane is

(Length of the incline) / (its height)

= (10m) / (1m)

= 10 .

It's the same for any load, and doesn't depend on the mass that you're trying to move up or down the ramp.

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A car starts from rest at a stop light and reaches 20m/s in 3.5s. Determine the acceleration of the car

olga55 [171]

The car accelerated at around ~5.7 m/s

8 0

3 years ago

Four forces act on a hot-air balloon, as shown

dolphi86 [110]

The magnitude of the resultant force on the balloon is 374.13 N.

The given forces from the image;

Upward force = 514 N
Downward force = 267 N
Eastward force = 678 N
Westward force = 397 N

The net vertical force on the balloon is calculated as follows;

$F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N$

The net horizontal force on the balloon is calculated as follows;

$F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N$

The magnitude of the resultant force on the balloon is calculated as follows;

$F = \sqrt{F_y^2 + F_x^2} \\\\F = \sqrt{(247)^2 + (281)^2} \\\\F= 374.13 \ N$

Thus, the magnitude of the resultant force on the balloon is 374.13 N.

Learn more here:brainly.com/question/4404327

5 0

2 years ago

If an object is projected upward with an initial velocity of 127 ft per? sec, its height h after t seconds is h equals negative

Stolb23 [73]

To determine the height of the object given the time, we simply use the given relation between height and time in the problem statement. It is given as:

h = -16t^2 + 127t

We substitute 55 seconds to t and obtain,

h = -16(55)^2 + 127(55)
h = - 41415

4 0

3 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago