Ask question

Ask question

All categories

faltersainse [42]

2 years ago

13

A horizontal uniform bar of mass 2.5 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

he end of the bar, and string 2 is attached a distance 0.6 m from the other end. A monkey of mass 1.25 kg walks from one end of the bar to the other.
Required:
Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.

1 answer:

11Alexandr11 [23.1K]2 years ago

6 0

Answer: I am not sure if you wanted me to answer this or not.

Explanation:

You might be interested in

A bird flies at a speed of 2.3 m/s if it has 14 j of kinetic energy what is the mass

Sidana [21]

Kinetic Energy = 1/2 * mv²

Kinetic Energy = 14 J, v = 2.3 m/s , m = ?

14 = 1/2 * m* 2.3²

14 = 0.5*m*2.3*2.3

m = 14 / (0.5*2.3*2.3)

m = 5.29 kg.

Mass = 5.29 kg.

7 0

3 years ago

Read 2 more answers

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago

A box of unknown mass is sliding with an initial speed vi = 4.70 m/s across a horizontal frictionless warehouse floor when it en

malfutka [58]

Here Change in Kinetic Energy = Work Done by Friction

Therefore, substituting the given values to the equation, we get

0.5 * m * (vFinal^2 - vInitial^2) = µ m g * d

Therefore

0.5*( 5.90^2 - Vfinal^2 ) = 0.100*9.8*2.10

Therefore

vfinal = 5.54 m/sec

<span> </span>

8 0

3 years ago

Which best describes the ages of the rocks on opposite sides of and the same distance from, A Seafloor spreading center?

Genrish500 [490]

Answer;

-The rocks are the same age

Explanation;

Seafloor spreading is the process by which the seafloor moves apart at mid-ocean ridges. Divergent seafloor spreading occurs at this type of plate boundary.

Seafloor spreading and other tectonic activity processes are the result of mantle convection. Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle’s convection currents makes the crust more plastic and less dense.

4 0

3 years ago

Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

3 years ago