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3 years ago

6

URGENT: Two horizontal forces act on an object on a surface. There is a 20 Newton applied force to the left and a sliding fricti

onal force of 5 Newtons to the right.
What is the net force on the object?

1 answer:

anygoal [31]3 years ago

7 0

Answer:

xbzcfhkuvy

Explanation:

dvdvxvdzdx

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Which season is signaled by average lower temperature and indirect, angled sunlight?

anygoal [31]

When the earth in its translation movement has the earth's axis is further from the sun and the energy per unit area is minimal, we have the winter season

The translation is the movement of the earth around the Sun, this movement establishes the duration of the year, also the earth's axis is inclined about 23.5º in relation to the orbit.

A consequence of this inclination of the Earth's axis is that the solar rays reach the surface of the planet with different inclinations, when they arrive more perpendicularly the energy deposited by the surface unit is greater and therefore the temperature increases, this period is called Summer.

When the rays arrive with the greater inclination, the energy deposited per unit area is minimal, which is why the temperature decreases, this period is called Winter.

In the periods when the axis is almost vertical we have an average absorption of energy, these two periods are called Spring and Autumn.

In conclusion, the winter season occurs when the terrestrial axis is furthest from the sun and the energy per unit of area is minimal.

Learn more about the seasons here:

brainly.com/question/20435793

4 0

2 years ago

WILL MARK BRAIN<br> A=154<br> B=145<br> C=26<br> D=206

ANEK [815]

Answer:

154

Explanation:

5 0

3 years ago

What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

3 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

<em>weight of the load, w = 557 N</em>
<em>force applied , F = 410 N</em>
<em>angle of force, = 15°</em>
<em>coefficient of kinetic friction = 0.46</em>
<em>distance moved, d = 6.5 m</em>

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

<em>m is the mass of the recycling bin</em>
<em /> $F_k$ <em> is the frictional force </em>

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

2 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

$F_{racquet}=-2F_{player}$

Hence the magnitude of Force by racquet is twice the Force by player

5 0

3 years ago