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AleksAgata [21]
3 years ago
5

Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. The cubes are lined up

from left to right and a force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2 . What is the magnitude of the force exerted on the middle cube by the left cube in this case?

Physics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

24 N

Explanation:

m = mass of the cube = 6.0 kg

Consider the three cubes together as one.

M = mass of the three cubes together = 3 m = 3 (6.0) = 18 kg

a = acceleration of the combination = 2 ms⁻²

F = Force applied on the combination

Using Newton's second law

F = ma = (18) (2) = 36 N

F_{L} = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N

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A bus starts from rest.if the acceleration is 2m/s square, find
MrMuchimi

Answer:

The velocity after 2 seconds can be found through:

V = u +a*t

Where V is final velocity, u is initial velocity, a is acceleration and t is time.

V = 0 + 2* 2= 4 meters/second

The distance (s) can be found through:

V^2= u^2 +2*a* s

Where V is final velocity, u is initial velocity, a is acceleration.

4^2= 0^2 + 2 *2*s

16= 0 + 4s

s= 4 meters

Distance (s) can also be found through:

s= ut + 1/2 at^2

s= 0+ 1/2 *2*2^2= 1 *2*2

s= 4 meters

Explanation:

3 0
2 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo
irinina [24]

Answer:

The torque about the origin is 2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}

Explanation:

Torque \overrightarrow{\tau} is the cross  product between force \overrightarrow{F} and vector position \overrightarrow{r} respect a fixed point (in our case the origin):

\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]

\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})

\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})

\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}

4 0
3 years ago
How to find direction of magnetic field.
Tju [1.3M]

unlock your third eye and youll be able to see it

6 0
2 years ago
A uniform 1.6-kg rod that is 0.89 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs han
artcher [175]

Answer:

8.27°

Explanation:

To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.

d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m

by a simple trigonometric relation you obtain that the angle:

sin\theta=\frac{d}{l}=\frac{0.128m}{0.89m}=0.144\\\\\theta=sin^{-1}(0.144)=8.27\°

hence, the angle between the rod and the horizontal is 8.27°

4 0
3 years ago
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