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3 years ago

5

An object of mass m1= 5kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and the

n is fastened to a hanging object of mass m2= 8kg as shown in figure.
Calculate,
(a) the acceleration (a) of the objects
(b) the tension (T) in the string.

(a) The acceleration of the objects (in m/s2) is

Answer for part 1
(b) The tension in the string (in Newtons) is

Answer for part 2

1 answer:

horrorfan [7]3 years ago

5 0

Answer:

a. 6.04 m/s^2 (3 sig. fig.) b. T = 30.2 N

Explanation:

a) Assume that the acceleration due to gravity is 9.81 m/s^2 downwards

8x9.81 = (5+8)a

a = 6.04 m/s^2 (3 sig. fig.)

b) Using the results from (a)

T = 5(6.04)

T = 30.2 N

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3. Convert 588 km/hr to m/s.

Luden [163]

Answer:

= 163.3 m/sec

hope it helps

7 0

3 years ago

To determine the types of equipment your boat must carry, you need to make a measurement of the boat. What is this measurement?

jenyasd209 [6]

Explanation:

In total, the length is measured from the tip of the bow in a linear fashion to the stern of the formation of delight including any back-deck extensions. The measurement involves bow sprits; rudders; detachable engines and engine sections; handles; and various fittings and connections.

Importance in calculating a boat's length:

it affects the transportation costs (the longer the length, the higher the cost). The pontoon's length counts as you find out how much rope you need to wrestle.

The cost of vessel settlement on marinas depends in part on the pontoon length. As more area is consumed by a more drawn pontoon, the docking charges are higher.

Transportation guidelines will probably not allow pontoons past a specific length on specific occasions of the day.

6 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped

Akimi4 [234]

Answer: F = 102141N

Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

$a=\frac{\Delta v}{\Delta t}$

$a=\frac{11}{0.14}$

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>

6 0

2 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

3 years ago