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A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

When a comet comes close to the Sun, what happens to it that makes it bright and easier to see?

Eddi Din [679]

Answer:

Explanation:

Comet is made by dust particles, icy particles, gases etc.

A comet has a fixed time to complete a revolution around the sun.

As a comet comes nearer to the sun, due to the heat of the sun the vapour and the icy particles becomes gases and due to the radial pressure of energy od sun, we observe a tail of comet which has vapours mainly. SI the comet is visible easily.

7 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

distance travel in the reaction time

d₁ = v x t

d₁ = 20 x 0.5 = 10 m

distance travel after you apply brake

using equation of motion

v² = u² + 2 a s

0 = 20² - 2 x 10 x s

s = 20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

distance between car and the deer = 38 m - 30 m

= 8 m

b) now, maximum speed car.

distance travel in reaction time

d₁ = s x t

d₁ = 0.5 V

distance left between them

d₂ = 38 - d₁

d₂ = 38 - 0.5 V

distance travel after you apply brake

using equation of motion

v² = u² + 2 a d₂

0 = (V)² - 2 x 10 x (38 - 0.5 V)

V² + 10 V - 760 = 0

now, solving the quadratic equation

$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}$

V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

7 0

3 years ago

Consider two diffraction gratings with the same slit separation. The only difference between the two gratings is that one gratin

kobusy [5.1K]

Answer:

True The grid with more slits gives more angle separation increases

True. The grating with 10 slits produces better-defined (narrower) peaks

Explanation:

Such a system can be seen as a diffraction network in this case with different number of lines per unit length, the expression for the constructive interference of a diffraction network is

d sin θ = m λ

where d is the distance between slits or lines, m the order of diffraction and λ the wavelength.

For network with 5 slits

d = 1/5 = 0.2

For the network with 10 slits

d = 1/10 = 0.1

let's calculate the separation (teat) for each one

θ = sin⁻¹ (m λ / d)

for 5 slits

θ₅ = sin⁻¹ (m λ 5)

for 10 slits

θ₁₀ = sin⁻¹ (m λ 10)

we can appreciate that for more slits the angle increases

the intensity of a series of slits is

I = I₀ sin²2 (N d/2) / sin² d/2)

when there are more slits (N) the peaks have greater intensity and are more acute (half width decreases)

let's analyze the claims

False

True The grid with more slits gives more angle separation increases

False

True The expression for the intensity of the diffraction peaks the intensity of the peaks increases with the number of slits as well as their spectral width decreases

False

5 0

3 years ago

A brick and a feather fall to the earth at their respective terminal velocities. Which objectexperiences the greater force of ai

Karo-lina-s [1.5K]

Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero. Thus, neither object is accelerating.

Recall Newton's second law: $\sum {\vec {F}}=m \vec {a}$

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, $F_{gravity}=mg$ , so the Force of Gravity acting on an object is directly proportional to the object's mass. (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

4 0

2 years ago