Help me please!!!!!!!!!!!!!/ Science

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

3 years ago

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip

Murrr4er [49]

Answer:

Explanation:

Given that the weight of the crate is

M=48kg

Then, weight of boy is

W=mg=48×9.8=470.4N

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

N=W

The normal equals weight of boy

N=470.4N

Then, applying frictional force opposing the boy motion.

Fr=μN

Fr=0.5×470.4

Fr=235.2N

Then, this is the forward force that the boy try using to pull the crate.

Since he can barely move his foot he has not yet overcome the coefficient of static friction.

ΣF = ma. , along x-axis

a along x-axis is 0 since he cannot move his foot, I.e he was not moving

F-Fr= 0

F=Fr=235.2N

The forward force the boy apply is 235.2N on the crate

Also, analysing the crate.

The forward force is now F=235.2N

Then the frictional force =Fr

Then,

ΣF = ma. , along x-axis

a along x-axis is 0 since the crate did not move,

F-Fr=0

Fr=F=235.2N

This is the frictional force on the crate.

Then using frictional law

Fr=μN

N=Fr/μ

N=235.2/0.9

N=261.33N

Now this normal is equal to the weight of the crate

ΣF = ma ,Along y axis

Since the body is not moving in y direction then, a=0 along y axis

N-W=0

W=N=261.33N

Then, since weight is given as

W=mg

m=W/g

m=261.33/9.81

m=26.64kg

The mass of the crate is 26.64kg

7 0

3 years ago

Read 2 more answers

The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a

Setler [38]

Answer:

x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

$x_{cm}$ = 1 / $M_{total}$ ∑ $x_{i} m_{i}$

Where M_{total} is the total masses of the system, $x_{i}$ is the distance between the particles and $m_{i}$ is the masses of each body

Let's apply this equation to our problem

M = Me + m

M = 5.98 10²⁴ + 7.36 10²²

M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

x_{cm} = 1 / 605.36 10²² [Me 0 + 7.36 10²² 3.82 10⁸]

x_{cm} = 4.644 10⁶ m

4 0

3 years ago

A train stops suddenly at the train station and everyone in the cars continue moving forward.

IgorC [24]

Answer:

nope

Explanation:

5 0

3 years ago

A uniform circular disk of moment of inertia 8.0 kg.m² is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the d

Mice21 [21]

Answer:

$\omega'=32\ rad.s^{-1}$

Explanation:

Given:

moment of inertial of a uniform circular disk, $I=8\ kg.m^2$

angular speed of rotation, $\omega=4\ rad.s^{-1}$

Mass of lump, $m'=1\ kg$

position of the lump from the center of the disk, $r'=1\ m$

<u>Using the conservation of the angular momentum:</u>

$I.\omega=I'.\omega'$

$8\times 4=m'.r'^2\times \omega'$

$32=1\times 1\times\omega'$

$\omega'=32\ rad.s^{-1}$

4 0

3 years ago