Ask question

Ask question

All categories

2 years ago

12

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.

You might be interested in

Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.

dimulka [17.4K]

Answer:

The distance of separation is $d = 0.092 \ m$

Explanation:

The mass of the each ball is $m= 10 g = 0.01 \ kg$

The negative charge on each ball is $q_1 =q_2=q = 1 \mu C = 1 *10^{-6} \ C$

Now we are told that the lower ball is restrained from moving this implies that the net force acting on it is zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

$F = \frac{kq_1 * q_2}{d}$

=> $m* g = \frac{kq_1 * q_2}{d}$

here k the the coulomb's constant with a value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

So

$0.01 * 9.8 = \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}$

$d = 0.092 \ m$

5 0

2 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

kobusy [5.1K]

Answer:

σ₁ = $3.167 * 10^{-6}$ C/m²

σ₂ = $7.6 * 10 ^{-6}$ C/m²

Explanation:

The given data :-

i) The radius of smaller sphere ( r ) = 5 cm.

ii) The radius of larger sphere ( R ) = 12 cm.

iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m

$E_{1} = (\frac{1}{4\pi\epsilon }) (\frac{Q_{1} }{R^{2} } )$

$358000 = 9 * 10^{9 } *\frac{Q_{1} }{0.12^{2} }$

Q₁ = 572.8 $* 10^{-9}$ C

Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.

V = constant

∴ $\frac{Q_{1} }{R} = \frac{Q_{2} }{r}$

$Q_{2} = \frac{r}{R} *Q_{1}$

$Q_{2} = \frac{5}{12} *572.8*10^{-9}$ = $238.666 *10^{-9}$ C

Surface charge density ( σ₁ ) for large sphere.

Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².

σ₁ = $\frac{Q_{1} }{A_{1} }$ = $\frac{572.8 *10^{-9} }{0.180864}$ = $3.167 * 10^{-6}$ C/m².

Surface charge density ( σ₂ ) for smaller sphere.

Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².

σ₂ = $\frac{Q_{2} }{A_{2} }$ = $\frac{238.66 *10^{-9} }{0.0314}$ = $7.6 * 10 ^{-6}$ C/m²

8 0

2 years ago

A boy on a bicycle approaches a brick wall as he sounds his horn at a frequency 400 hz. the sound he hears reflected back from t

Mashutka [201]

As the question is about changing in frequency of a wave for an observer who is moving relative to the wave source, the concept that should come to our minds is "Doppler's effect."

Now the general formula of the Doppler's effect is:
$f = (\frac{g + v_{r}}{g + v_{s}})f_{o}$ -- (A)

Note: We do not need to worry about the signs, as everything is moving towards each other. If something/somebody were moving away, we would have the negative sign. However, in this problem it is not the issue.

Where,
g = Speed of sound = 340m/s.
$v_{r}$ = Velocity of the receiver/observer relative to the medium = ?.
$v_{s}$ = Velocity of the source with respect to medium = 0 m/s.
$f_{o}$ = Frequency emitted from source = 400 Hz.
$f$ = Observed frequency = 408Hz.

Plug-in the above values in the equation (A), you would get:

$408 = ( \frac{340 + v_{r}}{340 + 0})*400$

$\frac{408}{400} = \frac{340 + v_{r}}{340}$

Solving above would give you,
$v_{r}$ = 6.8 m/s

The correct answer = 6.8m/s

7 0

2 years ago

A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen

Svetllana [295]

Answer:

$h_p = 30.46\ m$

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

$v_f=g\cdot t$

Where $g = 9.8 m/s^2$

Similarly, the distance y the object has traveled is calculated as follows:

$\displaystyle y=\frac{g\cdot t^2}{2}$

If we know the height h from which the object was dropped, we can solve the above equation for t:

$\displaystyle t=\sqrt{\frac{2\cdot y}{g}}$

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

$\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec$

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

$t_2=2.56 - 2 = 0.56\ sec$

Therefore, it has traveled down a distance:

$\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m$

Thus, the height of the pen is:

$h_p = 32 - 1.54\Rightarrow h_p=30.46\ m$

8 0

2 years ago

One end of a rope is tied to the handle of a horizontally-oriented and uniform door. a force fis applied to the other end of the

sergeinik [125]

<span>Answer:The weight of the door creates a CCW torque given by Tccw = 145 N*3.13 m / 2 You need a CW torque that's equal to that Tcw = F*2.5 m*sin20</span>

4 0

3 years ago

Read 2 more answers