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DedPeter [7]
3 years ago
5

Truck drivers probably cannot see your vehicle if____.

Physics
1 answer:
laiz [17]3 years ago
6 0
Answer Is Tailgating
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1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increa
ser-zykov [4K]

(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

<em>Let the </em><em>angle</em><em> of inclination of the </em><em>ramp</em><em> = θ</em>

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

F_n = mgcos (\theta)

when θ is 0;

F_n = mgcos (0)\\\\F_n = mg

when θ is 90;

F_n = mgcos(90)\\\\F_n = 0

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

F_x = mgsin(\theta)\\\\

when θ is 0;

F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0

when θ is 90;

F_x = mgsin(90)\\\\F_x = mg

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

F_n = \mu F_n

F_k = \mu mgcos(\theta)

F_k = \mu mg cos(0)\\\\F_k = \mu mg

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N

Learn more here: brainly.com/question/14121363

7 0
3 years ago
NEED HELP THIS IS DUE IN 30 MINUTES
11111nata11111 [884]

-- The total distance between the towns = 460 km.

-- For the first 3 hours, he travels at an average speed of 60 km/hr.  During that time, he covers (60 km/hr) x (3 hrs) = 180 km.

-- At that point, he still has (460km - 180km) = 280 km to go.

-- If he drives that 280km at an average speed of 70 km/hr, it'll take him (280km)/(70km/hr) = 4 hrs to cover it.

-- His total time for the whole trip is (3 hours for the first part) + (4 hours for the last part) = <em>7 hours</em>.

8 0
3 years ago
A string is attached to a ball that has a mass of 0.11 kg. A student pulls up on the string so that the ball accelerates upward
enot [183]

Answer:

T=+1.133N

Explanation:

Tension and weight are forces that have opposite directions

Weight is negative (downward)

W=m*g= 0.11kg*(-9.8m/s^2)

W= -1.078N

Tension is possitive (upward)

The total force will be the sum of both (the difference taking in consideration the direction)

Ft= T+W

Also the total force is the product of the mass due to acceleration:

Ft=m*a

Ft= +0.11kg*0.5m/s^2

Ft=+0.055N (upward)

Tension will be the difference between Ft and W:

T= Ft-W

T=+0.055N-(-1.078N)

T=+1.133N

7 0
3 years ago
Physics and chemistry is the study of what? ​
LenKa [72]
<h3><em>physical</em><em> </em><em>science</em><em> </em><em>deals</em><em> </em><em>with</em><em> </em><em>the</em><em> </em><em>study</em><em> </em><em>of</em><em> </em><em>physics</em><em> </em><em>chemistry</em></h3>

Explanation:

yan lng po Alam ko

8 0
2 years ago
A crude approximation for the x component of velocity in an incompressible laminar boundary layer is a linear variation from u =
slega [8]

Answer:

2.5 * 10^-3

Explanation:

<u>solution:</u>

The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:

<em>δu/δx+δv/δy=0</em>

so that:  

<em>δv/δy= -δu/δx</em>

Now, since u = Uy/δ, where δ = cx^1/2, we have that:

<em>u=U*y/cx^1/2</em>

and we obtain:  

<em>δv/δy=U*y/2cx^3/2</em>

The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):  

v=∫δv/δy(dy)=U*y/4cx^1/2

 =y/x*(U*y/4cx^1/2)

 =u*y/4x

which is exactly what we needed to demonstrate.  

Also, using u = U*y/δ in the last equation we can obtain:  

v/U=u*y/4*U*x

     =y^2/4*δ*x

which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:

(v/U)_max=δ^2/4δx

                =δ/4x

                =2.5 * 10^-3

7 0
3 years ago
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