Ask question

Ask question

All categories

3 years ago

14

Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 25°, and Paul is pulling a large crate u

p the ramp with a rope that angles 10° above the ramp. If Paul pulls with a force of 550 N, what are the horizontal and vertical components of his force? (Force is measured in newtons, abbreviated N.)

1 answer:

Alisiya [41]3 years ago

4 0

Explanation:

It is given that,

The ramp is tilted upwards at 25 degrees and Paul is pulling a large crate up the ramp with a rope that angles 10° above the ramp.

Total angle with respect to ramp is 35 degrees.

If Paul pulls with a force of 550 N.

The horizontal component of the force is given by :

$F_x=F\ cos\theta$

$F_x=550\ cos(35)$

$F_x=450.53\ N$

The vertical component of the force is given by :

$F_y=F\ sin\theta$

$F_y=550\ sin(35)$

$F_y=315.46\ N$

Hence, this is the required solution.

You might be interested in

You drive a car east on the highway at 26 m/s. Another car passes you moving east traveling at 32 m/s. How fast do you view the

tamaranim1 [39]

The speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

<h3>Relative velocity of the car</h3>

The speed of the car passing you is determined by applying relative velocity principle as shown below;

Vr = Va - Vb

Vr = 26 m/s - 32 m/s

Vr = -6 m/s

Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.

Learn more about relative velocity here: brainly.com/question/17228388

#SPJ1

3 0

2 years ago

The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m

butalik [34]

$\sf\underline{Solution:}$

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

$\sf{Now,according \: to \:the\:question:}$

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

<u>☆</u><u> </u><u>Substitute</u><u> </u><u>the</u><u> </u><u>required</u><u> </u><u>values</u><u> </u><u>in</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>formula</u><u>-</u>

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0

2 years ago

A 90 kg astronaut Travis is stranded in space at a point 12 m from his spaceship. In order to get back to his ship, Travis throw

insens350 [35]

Answer:

Explanation:

This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:

$[m_av_a+m_ev_e]_b=[m_av_a+m_ev_e]_a$ which, in words, is

The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:

$[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a$

Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).

0 = 90.0v - 2.0 and

2.0 = 90.0v so

v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.

12 = .022t and

t = 550 seconds, which is the same thing as 9.2 minutes

7 0

3 years ago

You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizonta

Citrus2011 [14]

Answer:

$\triangle x=3.2*10^-^5 m$

Explanation:

From the question we are told that

Height of circular cylinder is $h= 1.8m$

Diameter of cylinder $D=4cm=>0.04m$

Horizontal Force $F=900N$

$Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\$

Generally the formula for shear modulus is mathematically represented by

$G=\frac{\tau}{\gamma}$

Where

$G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain$

$G=\frac{F/A}{\triangle x/L}$

$G=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900}{\pi r^2} *\frac{1.8}{\triangle x}$

${\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}$

$\triangle x=3.2*10^-^5 m$

5 0

2 years ago

A car applies a force of 20,000 N downward towards the street.

NikAS [45]

Answer:

The street applies a force of 20,000 N on the car.

Explanation:

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. The car applies the same amount of force to the street as the street to the car.

5 0

3 years ago