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2 years ago

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A horse and rider are racing to the right with a speed of 21\,\dfrac{\text m}{\text s}21 s m 21, start fraction, start text, m

, end text, divided by, start text, s, end text, end fraction when they pass the finish line and begin slowing down. The horse slows for 52\,\text m52m52, start text, m, end text with constant acceleration before it stops. What was the acceleration of the horse as it came to a stop? Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

2 answers:

abruzzese [7]2 years ago

8 0

Answer:

The acceleration of the horse as it came to a stop is $4.24\ m/s^2$ .

Explanation:

Given that,

Initial speed of the horse and rider, u = 21 m/s

It finally comes to rest, v = 0

The horse slows down for 52 m with constant acceleration before it stops.

We need to find the acceleration of the horse as it came to a stop. Let a is the acceleration. We can find it using third equation of motion as :

We need to find the acceleration of the horse as it came to a stop. Let a is the acceleration. We can find it using third equation of motion as :

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{-u^2}{2d}\\\\a=\dfrac{-(21)^2}{2\times 52}\\\\a=-4.24\ m/s^2$

So, the acceleration of the horse as it came to a stop is $4.24\ m/s^2$ . Negative sign shows deceleration.

Marrrta [24]2 years ago

7 0

Answer:

a=-4.2 m/s²

Explanation:

The horse riding so inital velocity is given finally the rider stops so the final velocity is zero.

initial velocity =Vi= 21 m/s

final velocity =Vf= 0 m/s

distance covered = S=52 m

By using 2nd equation of motion we can find the acceleration

2aS=Vf² -Vi²

a=(-441)/104

a=-4.2 m/s²

So the accceleration is 4.2 m/s².

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

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2 years ago

A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

vekshin1

Electric charge on the plastic cube: $1.3\cdot 10^{-7}C$

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

$V(r)=\frac{kQ}{r}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

$Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V$

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

$qV' = \frac{1}{2}mv^2$

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

$q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C$

Learn more about electric fields:

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2 years ago

luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that

Andreas93 [3]

Answer:

a) W = 643.5 J, b) W = -427.4 J

Explanation:

a) Work is defined by

W = F. x = F x cos θ

in this case they ask us for the work done by the external force F = 165 N parallel to the ramp, therefore the angle between this force and the displacement is zero

W = F x

let's calculate

W = 165 3.9

W = 643.5 J

b) the work of the gravitational force, which is the weight of the body, in ramp problems the coordinate system is one axis parallel to the plane and the other perpendicular, let's use trigonometry to decompose the weight in these two axes

sin θ = Wₓ / W

cos θ = Wy / W

Wₓ = W sinθ = mg sin θ

Wy = W cos θ

the work carried out by each of these components is even Wₓ, it has to be antiparallel to the displacement, so the angle is zero

W = Wₓ x cos 180

W = - mg sin 34 x

let's calculate

W = -20 9.8 sin 34 3.9

W = -427.4 J

The work done by the component perpendicular to the plane is ero because the angle between the displacement and the weight component is 90º, so the cosine is zero.

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2 years ago

An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni

Marysya12 [62]

When air is blown into the open pipe,

L = $\frac{nλ}{2}$

where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation

⇒λ= $\frac{2L} {n}$

Note here that n=1 is for fundamental, n=2 is first harmonic and so on..

⇒ third harmonic will be n=4

Given L=6m, n=4, solving for λ we get:

λ= $\frac{(2)*(6)}{4}$ =3m

Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:

c=f.λ Or f= $\frac{c}{λ}$

⇒f= $\frac{344}{3}$

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2 years ago

Experiments using "optical tweezers" measure the elasticity of individual DNA molecules. For small enough changes in length, the

GalinKa [24]

Answer:

Spring constant, k = 0.3 N/m

Explanation:

It is given that,

Force acting on DNA molecule, $F=1.5\ nN=1.5\times 10^{-9}\ N$

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Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :

$F=-kx$

$k=\dfrac{F}{x}$

$k=\dfrac{1.5\times 10^{-9}\ N}{5\times 10^{-9}\ m}$

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8 0

2 years ago