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goldfiish [28.3K]

2 years ago

4

A rightward force is applied to a 10-kg object to move it across a rough surface at constant velocity. The coefficient of fricti

on between the object and the surface is 0.2. Use the diagram to determine the (A) gravitational force, (B) normal force, (C) applied force, (D) frictional force, and (E) net force. (Neglect air resistance.)

1 answer:

Alinara [238K]2 years ago

8 0

Answer:

Explanation:

Gravitational force = normal force which are both the same thing as the weight of the object which is found in w = mg. Therefore,

w = 10(9.8) = 98 N.

If the object is being pulled at a constant velocity, then the acceleration of said object is 0. Thus,

F - f = ma and

F - f = 0 so

F = f. So we just need to solve for f, which is found in the equation

f = μ $F_n$ so

f = .2(98) and

f = 19.6 = F

The net force is 0 if F = f

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Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as

Sonja [21]

Explanation:

The waveform expression is given by :

$y(x,t)=y_m\ sin(0.333x+5.36+585t)$ ...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

$y(x,t)=y_m\ sin(kx+\phi+\omega t)$ ..........(2)

Where

$k=\dfrac{2\pi}{\lambda}$

$\omega=2\pi f$

On comparing equation (1) and (2) we get :

$0.333=\dfrac{2\pi}{\lambda}$

$\lambda=18.86\ m$

$585=2\pi f$

f = 93.10 Hz

Time period, $T=\dfrac{1}{f}$

$T=\dfrac{1}{0.010}$

T = 0.010 s

Phase constant, $\phi=5.36\ radian$

Hence, this is the required solution.

8 0

2 years ago

What makes the subject of star formation so difficult and complex? Star formation is too expensive to study in detail. Stars liv

marta [7]

Answer:

Stars live to long to be observed from birth to death

Explanation:

The life expectancy of a star are millions of years compared to that of human thereby making it difficult to study the life path and activities of the stars till the point of extinction. This brought about scientist studying different star with different ages in order to answer their burning question on formation of stars which is called the figuring out process(stellar activity) and this makes it difficult and tedious.

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2 years ago

A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

2 years ago

Assume you are driving 20 mph on a straight road. Also, assume that at a speed of 20 miles per hour, it takes 100 feet to stop.

Viefleur [7K]

Answer:900 feet

Explanation:

Given

Velocity $\left ( V_1\right )=20 mph\approx 29.334 ft/s$

it take 100 feet to stop

Using Equation of motion

$v^2-u^2=2as$

where

v,u=Final and initial velocity

a=acceleration

s=distance moved

$0-\left ( 29.334\right )^2=2\left (-a\right )\left ( 100\right )$

$a=\frac{29.334^2}{2\times 100}=4.302 ft/s^2$

When velocity is 60 mph $\approx 88.002 ft/s$

$v^2-u^2=2as$

$0-\left ( 88.002\right )^2=2\left ( -4.302\right )\left ( s\right )$

s=900.08 feet

8 0

2 years ago

Read 2 more answers

If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t

Whitepunk [10]

From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer: 283 s (nearest second)

6 0

2 years ago