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Harlamova29_29 [7]

2 years ago

3

Three forces are exerted on an object placed on a slope in the figure. The forces are measured in newtons (N). 1. Assuming that

forces are vectors, what is the component of the net force F⃗ net=F⃗ 1+F⃗ 2+F⃗ 3 parallel to the slope? 2. What is the component of F⃗ net perpendicular to the slope? 3. What is the magnitude of F⃗ net? 4. What is the direction of F⃗ net?

1 answer:

ehidna [41]2 years ago

8 0

The component of the net force that is parallel to the slope is the friction force. It is the force due to the friction present between the object and the surface of the slope. The force perpendicular to the slope is called the normal force and it is the contact force exerted by the object to the surface of the slope.

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Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo

alexgriva [62]

Answer:

R=19.5m

$\theta$ = 4.65° S of W

Explanation:

Refer the attached fig.

displacement of the x and y components

x-component displacement is ( $R_{x}$ ) = $A_{x}+B_{x}$

= A $\sin$ (20°) + B $\cos$ (40°)

= -12.0 $\sin$ (20°) + 20.0 $\cos$ (40°)

= -19.425m

x-component displacement is ( $R_{y}$ ) = $A_{y}+ B_{y}$

= A $\cos$ (20°) - B $\sin$ (40°)

= 12.0 $\cos$ (20°) - 20.0 $\sin$ (40°)

= -1.579

resultant displacement

∴

R = $\sqrt{R_{x}^{2} +R_{y}^{2} } }$

= $\sqrt{(-19.425)^{2}+(-1.579)^{2} }$

=19.5m

$\theta$ = $\tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |$

$\theta$ = $\tan^{-1}\left | \frac{1.579}{19.425} \right |$

$\theta$ = 4.65° S of W

6 0

2 years ago

A. Why is the stratosphere considered a "stable" layer in the atmosphere?

vovangra [49]

Answer:

A. Stratosphere is said to be stable layer of the atmosphere when cool air sinks and warm air rises.Due to the fact that cool air has tendency to sink ,the air is not going fluctuating up and down in the stratosphere. This means that the air remains stationary or particles remains there for a very long duration.

B. If the lifted index is negative then the parcel temperature is warmer than the actual temperature. In addition, the parcel that is less warm than the surrounding will be less dense and will rise.

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D. the threshold used by storm chasers to assess if the dew point temperature is high enough to produce large thunderstorms is moisture ,the surface dew point needs to be 55 degrees fahrenheit or greater for a surface based thunderstorm to occur.

E. Wind shear is the change in wind direction or speed with height in an atmosphere.

Explanation:

7 0

2 years ago

A 40-kg box is being pushed along a horizontal smooth surface. The pushing force is 15 n directed at an angle of 15Â° below the

Kobotan [32]

Answer:

Acceleration of the crate is 0.362 m/s^2.

Explanation:

Given:

Mass of the box, m = 40 kg

Applied force, F = 15 N

Angle at which the force is applied, $(\theta)$ = 15°

We have to find the magnitude of the acceleration.

Let the acceleration be "a".

FBD is attached with where we can see the horizontal and vertical component of force.

⇒ $F_x=Fcos(\theta)$ and ⇒ $F_y=Fsin (\theta)$

⇒ $F_x=15cos(15)$ ⇒ $F_y=15sin (15)$

⇒ Applying concept of forces.

⇒ $\sum F_x=F_n_e_t =F-f$

⇒ $F_n_e_t =F-f$

⇒ $ma =F-f$ <em> ...Newtons second law Fnet = ma</em>

⇒ $a =\frac{F-f}{m}$

⇒ Plugging the values.

⇒ $a =\frac{15cos(15)-0}{40}$ <em>...f is the friction which is zero here.</em>

⇒ $a =\frac{14.48}{40}$

⇒ $a=0.362\ ms^-^2$

Magnitude of the acceleration of the crate is 0.362 m/s^2.

4 0

2 years ago

At 25 °C, a bottle contains 2.00 L of water in its liquid state. What is the volume of the water after it freezes (at 0 °C)? The

Alex73 [517]

Answer:

2.175 L

Explanation:

temperature (T) = 25 degrees

volume of water (v) = 2 L = 2000 mL

density of water = 0.997 g/mL

density of ice = 0.917 g/mL

we can get the mass of the water and the use it to get the volume when it freezes to ice, this is because the mass remains the same irrespective of the change of state.

mass of water = volume x density = 2000 x 0.997 = 1994 g

volume of ice = mass/density = 1994 / 0.917 = 2174.5 mL = 2.175 L

6 0

2 years ago

In an experiment to illustrate the propagation of sound through fluids, identical sound sources emitting at 440 Hz are used to p

earnstyle [38]

Answer:

B. In mercury, the frequency of the wave is the same as in ethanol, but the wavelength is greater.

Explanation:

To solve this easily, we can just calculate the wavelength of the sound in Ethanol and in Mercury.

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λ = c/f

λ = 1160/440

λ = 2.63 m

In Mercury, the wavelength will be:

λ = c/f

λ = 1450/440

λ = 3.3 m

The wavelength of sound is greater in Mercury than in Ethanol but the frequency is the same.

Frequency of sound is not dependent on medium, but velocity and wavelength change depending on the medium.

3 0

2 years ago