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Natasha_Volkova [10]
2 years ago
6

Jonathan and Cody’s older brother Josh, who is pictured in the Figure below, is standing at the top of a half-pipe at Newton’s S

kate Park. Gravity is exerting a downward force on the skateboard as seen in the picture. Why doesn’t it tip over the edge and start rolling down the side of the half-pipe?
Physics
1 answer:
Margarita [4]2 years ago
8 0
In order to begin moving a Mass(Accelaration) will solve for the force required to move the skateboard
Mass(9.8) will give you the force needed to move the skateboard originally
You might be interested in
When a driver presses the brake pedal, his car can stop with an acceleration of -5.4m/s^2. How far will the car travel while com
Dahasolnce [82]
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m
3 0
3 years ago
Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that it
jek_recluse [69]

Answer:

a)    I= I₀ (cos²θ - cos⁴θ)    b) 75.5º

Explanation:

a) For this exercise we must use Malus's law

         I = I₀ cos² θ

where tea is the angle between the two polarizers.

We apply this expression to our case

* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical

         I₁ = I₀ cos² θ

* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.

The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical

              θ₂ = 90- θ’ = θ

fiate that in the exercise we must take a reference system and measure everything with respect to this system.

          I = I₁ cos² θ'

       

we substitute

         I = (I₀ cos² tea) cos² (θ - 90)

        cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ

         I = Io cos² θ sin² θ

        1= cos²θ+ sin²θ

       sin²θ = 1 - cos²θ

        I= I₀ (cos²θ - cos⁴θ)

b) to find when the intensity is maximum,

we can use that we have an extreme point when the drift is zero

          \frac{dI}{d \theta} = 0

          \frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0

whereby

            cos θ - 2 cos³ θ = 0

            cos θ ( 1 - 2 cos² θ) = 0  

The zeros of this function are in

           θ = 90º

           1-2cos²θ =0       cos θ = 0.25  θ =  75.5º

Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.

Consequently, the angle that allows the maximum intensity to pass is 75.5º

5 0
2 years ago
A coin is thrown with a velocity of 0 m/s down a dry well and hits bottom in 1.2s, what’s the depth of the well?
pentagon [3]

Answer:

The well is 7.1 meters deep.

Explanation:

The formula to use here is the distance in a uniformly accelerated motion:

d = \frac{1}{2}at^2+v_0t+d_0

where d stands for distance, t for time, a for acceleration, v0 and d0 for initial velocity and distance, respectively. Since the initial distance and velocity are both zero, we are left with the first term. The coin is in free fall and so it is accelerated by gravity:

d = \frac{1}{2}at^2= \frac{1}{2}gt^2=\frac{1}{2}9.8\frac{m}{s^2}1.2^2s^2=7.1m

The well is 7.1 meters deep.

5 0
4 years ago
You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The
Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is h=3.6\ m

Initial speed of Putty u=9.5\ m/s

Speed of Putty just before it strike the ceiling is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v^2-9.5^2=2\times (-9.8)\times 3.6

v^2=19.69

v=4.43\ m/s

time taken by putty to reach the ceiling

v=u+at

4.43=9.5-9.8\times t

t=\frac{5.07}{9.8}

t=0.517\ s

8 0
3 years ago
A train whose proper length is 1200 m passes at a high speed through a station whose platform measures 900 m, and the station ma
TiliK225 [7]

Answer:

0.66c

Explanation:

Use length contraction equation:

L = L₀ √(1 − (v²/c²))

where L is the contracted length,

L₀ is the length at 0 velocity,

v is the velocity,

and c is the speed of light.

900 = 1200 √(1 − (v²/c²))

3/4 = √(1 − (v²/c²))

9/16 = 1 − (v²/c²)

v²/c² = 7/16

v = ¼√7 c

v ≈ 0.66 c

6 0
3 years ago
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