Why Tractor have wide and big wheels

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower

liq [111]

initial speed of the stuntman is given as

$v = 28 m/s$

angle of inclination is given as

$\theta = 15 degree$

now the components of the velocity is given as

$v_x = 28 cos15 = 27.04 m/s$

$v_y = 28 sin15 = 7.25 m/s$

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

$\delta y = -25 m$

$\delta y = v_y * t + \frac{1}{2} at^2$

$-25 = 7.25 * t - \frac{1}{2}*9.8* t^2$

by solving above equation we have

$t = 3.12 s$

Now in the above interval of time the horizontal distance moved by it is given by

$d_x = v_x * t$

$d_x = 27.04 * 3.12 = 84.4 m$

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.

5 0

2 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$

=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

the solution is subtracted by 360° in order to get the actual angle

4 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m

mixas84 [53]

Answer:

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

U = $k \frac{q_1q_2}{r_{12}}$

for the proton at x = -1 m

U₁ = $- k \frac{e^2 }{r+1}$

for the electron at x = 1 m

U₂ = $k \frac{e^2 }{r-1}$

starting point.

Em₀ = K + U₁ + U₂

Em₀ = $\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}$

final point

Em_f = $k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})$

energy is conserved

Em₀ = Em_f

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})

\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( $\frac{2}{(r_2+1)(r_2-1)}$ )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ $- \frac{1}{20+1} + \frac{1}{20-1}$ ) = 9 109 (1.6 10-19) ²( $\frac{2}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( $\frac{1}{r_2^2 -1}$ )

2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷ $\frac{1}{r_2^2 -1}$

$\frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}$

r₂² -1 = (4.443 10⁸)⁻¹

r2 = $\sqrt{1 + 2.25 10^{-9}}$

r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0

3 years ago

A force of 720 Newton stretches a spring 4 meters. A mass of 45 Kilograms is attached to the spring and is initially released fr

poizon [28]

Answer:

x(t) = -3sin2t

Explanation:

Given that

Spring force of, W = 720 N

Extension of the spring, s = 4 m

Attached mass to the spring, m = 45 kg

Velocity of, v = 6 m/s

The proper calculation is attached via the image below.

Final solution is x(t) = -3.sin2t

5 0

3 years ago

Which of the following is a scalar quantity?

makvit [3.9K]

The only scalar quantity is a. 35 m

Explanation:

In physics, there are two types of quantities:

- Scalar: a scalar quantiy is a quantity having only a magnitude, so it is just a number followed by a unit. Examples of scalar quantities in physics are:

Speed

Energy

Distance

Time

- Vector: a vector quantity is quantity having both a magnitude and a direction. Examples of vector quantities in physics are:

Velocity

Force

Acceleration

Displacement

Let's now analyze each given option, to evaluate if it is a scalar or a vector:

a. 35m --> it is only a unit (no direction), so it is a scalar

b. 20 m to the right --> it has a direction, so it is a vector

c. 30 m to the North --> it has a direction, so it is a vector

So, the only correct option is a).

Learn more about scalars and vectors:

brainly.com/question/2678571

brainly.com/question/4945130

#LearnwithBrainly

4 0

2 years ago

Why Tractor have wide and big wheels​

Why Tractor have wide and big wheels