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andrew11 [14]
3 years ago
7

For a snowboard jumper in the air, what force or forces will be most important for modeling the motion?

Physics
1 answer:
Reptile [31]3 years ago
4 0

Answer: Gravitational force and drag force

Explanation:

For a snowboard jumper in the air, two forces would be acting. One in the downward direction- the gravitational pull and second in the opposite direction to the motion, the drag force due to air. If the snowboard jumper jumps in the air at a certain angle with the horizontal. The forces are written as the sum of vertical and horizontal components. Hence, for the modeling the motion, gravitational force and drag force are important,

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The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo
mylen [45]

You can just use basic trigonometry to solve for the x & y components.

<span>vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5></span>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

<span>vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

<span> </span>

6 0
3 years ago
I need help ASAP plzzzz
Fiesta28 [93]

Answer:

a) 1.75s b) 17.2 m/s (down)

Explanation:

d1= 15m d2= 0m (because it hits ground)

a= -9.81 m/s^2 t=???

Equation

the triangle means change in so d2-d1

Δd= v1 * t + 1/2 * a * t^2

0m-15m= v1*t + 1/2 a t^2

-15 m= 0m/s*t (goes away) + 1/2* a *t^2

-15mx2= t^2

-15mx2/a= t^2

Square root (-30/-9.81m/s^2)

t=1.75 s

b) now v2!!

Im going to use v2= v1 + a*t

v2= 0m/s + -9.81 x 1.75s

v2 = -17.2 m/s or you can say 17.2 m/s down!!!

7 0
3 years ago
Question 2
Yakvenalex [24]

Answer:

B

Explanation:

6 0
3 years ago
current is passed through two parallel conductors in the same direction. If the conductors are placed near each other,they will​
svlad2 [7]

If current is passed through two parallel conductors in the same direction and the conductors are placed near each other, they will​ attract each other.

<h3>What is electric current?</h3>

Electric current can be defined as the flow of electrons.

Since electrons are easily removed from atom and are very mobile, the flow of electrons constitute an electric current.

Materials which allow electric current to flow through them are known as conductors. Examples of conductors are metals, and electrolytes.

On the other hand, materials which do not allow electric current to pass through them are known as insulators. Examples of insulators are wood and rubber.

The flow of current is known as electricity.

Parallel conductors with current flowing through them in the same direction are attracted to each other as a result of a magnetic field produced by the flow of current.

In conclusion, conductors allow electric current to pass through and the flow of current through a conductor produces a magnetic field.

Learn more about parallel conductors at: brainly.com/question/17148082

#SPJ1

6 0
2 years ago
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
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