what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this

100 POINTS FOR CORRECT ANSWER/EXPLANATION

Shalnov [3]

Answer:

6 N

Explanation:

Let's start with the small block m on top. There are four forces:

Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).

Now let's look at the large block M on bottom. There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

Weight force Mg pulling down,

Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law. Sum of forces in the y direction for block m:

∑F = ma

N₁ − mg = 0

N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

F = gμ(3m + M)

Since M = 2m:

F = 5gμm

Plug in values:

F = 5 (10 m/s²) (0.400) (0.300 kg)

F = 6 N

8 0

3 years ago

Read 2 more answers

Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver

zalisa [80]

Answer:

Total time taken=110 seconds

Total distance traveled=480m

Explanation:

First of all, we find the total time taken:

For that, we use the formula : Distance/Speed= Time

Time for part 1 : 200/5=40 seconds

Time for part 2 : 280/4=70seconds

Total time taken=110 seconds

Total distance traveled=480m

Average Speed= 480/110=4.36 m/s

Total displacement=200-280=-80m (Since this is displacement, we need to find the distance between the initial and final point. Also, I've taken east direction as positive and west as negative)

Average Velocity=-80/110=-0.72 m/s

OR 0.72m/s towards west.

3 0

2 years ago

Barnard’s Star is a red dwarf. It is located 5.9 light years from Earth. (One light year is the same as 9.46 trillion kilometers

mote1985 [20]

A star is located 5.9 light years from Earth.
We know that : 1 light year = 9.46 trillion kilometers.
We will calculate the distance in trillion kilometers multiplying the number of light years by 9.46:
5.9 * 9.46 = 55.814
Answer: The distance is 55.814 trillion km.

5 0

3 years ago

Read 2 more answers

A wave traveling in water has a frequency of 500Hz and a wavelength of 3.00 meters. What is the speed of the wave?

jenyasd209 [6]

Answer:

1500 m/s

Explanation:

3 0

2 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