Ask question

Ask question

All categories

2 years ago

14

A 10kg crate sits at rest on a rough flat surface. Astudent decides to pull the crate by attaching a rope at a 37 degree angle.

Although the student pulls the rope with a force of 600 newtons, the coefficient of kinetic friction is large and has the force the student applies remains constant, how much time after he begins pulling the crate will it take before the crate has traveled a distance of 1.0 meter?

1 answer:

sertanlavr [38]2 years ago

5 0

Answer:

Explanation:

Normal force of the surface on the box will be

N = mg - Fsinθ

Ν = 10(9.8) - 600sin37

N = -263

As normal force cannot be less than zero, the applied force lifts the crate off the surface.

Now it's just a matter of finding the acceleration

In the horizontal direction, the acceleration is

a = F/m

a = (600cos37) / 10

a = 47.9181... m/s²

the crate weight is mg = 10(9.8) = 98 N.

In the vertical direction the acceleration is

a = ((600sin37 - 98) / 10)

a = 26.3089... m/s²

total acceleration is

a = √(47.9181² + 26.3089²)

a = 54.6653... m/s²

s = ½at²

t = √(2s/a)

t = √(2(1.0)/54.6653)

t = 0.19127...

t = 0.19 s

You might be interested in

Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

2 years ago

How long would it take for a ball dropped from the top of a 576-foot building to hit the ground? round your answer to two decima

LUCKY_DIMON [66]

If a ball is if a ball is dropped from a 576ft building it would take about 8 seconds for it to hit the ground.

4 0

2 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

The siren on an ambulance emits a sound of frequency 2.80×103Hz. If the ambulance is traveling at 26.0 m/s (93.6 km/h or 58.2 mi

Norma-Jean [14]

To solve this problem it is necessary to apply the concepts related to the described wavelength through frequency and speed. Mathematically it can be expressed as:

$\lambda = \frac{v}{f}$

Where,

$\lambda =$ Wavelength

f = Frequency

v = Velocity

Our values are given as,

$f = 2.8*10^3Hz$

$v = 340m/s \rightarrow$ Speed of sound

Keep in mind that we do not use the travel speed of the ambulance because we are in front of it. In case it approached or moved away we should use the concepts related to the Doppler effect:

Replacing we have,

$\lambda = \frac{340}{2.8*10^3}$

$\lambda = 0.1214m$

Therefore the frequency that you hear if you are standing in from of the ambulance is 0.1214m

5 0

2 years ago

In an electric header 500w is written. What is its meaning

Fiesta28 [93]

Answer:

Because it's wattage is 500W.

Explanation:

This is its mearurement

4 0

2 years ago