Ask question

Ask question

All categories

2 years ago

5

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 8 ft/s. Dete

rmine the velocity and acceleration of the hay bale when the horse is 10 ft away from the barn.

1 answer:

g100num [7]2 years ago

6 0

Answer:

Velocity = 8 ft/s

Acceleration = 0 m/s²

Explanation:

Since, the horse is moving with a constant velocity, whose magnitude is given as equal to 8 ft/s. Therefore, it will have the same velocity when it is 10 ft away from the barn. And the velocity of hay bale will be same as the velocity of horse, as the horse is carrying the bales. Therefore:

<u>Velocity = 8 ft/s</u>

Coming to the second part of the question, which relates to the acceleration of the hay bale, when horse is 10 ft away from the barn. The formula for acceleration is given as:

Acceleration = Change in Velocity/ Time

But, the velocity of the horse in constant, which means there is no change in velocity. Hence,

Change in Velocity = 0

Therefore,

Acceleration = 0/Time

<u>Acceleration 0 m/s²</u>

You might be interested in

You drop your keys in a high-speed elevator going up at a constant speed. Part APart complete Do the keys accelerate faster towa

anzhelika [568]

Answer:

Explained

Explanation:

a) No, the keys were initially moving upward in the elevator only effects the initial velocity of the key and not the rate of change of velocity that is acceleration. So, the keys accelerate with the same acceleration as before.

b)Yes, keys will accelerate towards the floor faster if it is a constant speed than it is moving downward because if the elevator is accelerating downward, the downward change in velocity of the keys is at least partially matched by a downward change in the velocity of the of the elevator.

5 0

2 years ago

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

2 years ago

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b

Butoxors [25]

Answer:

$v_{o}=18m/s$

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

$\beta =25\\y=2.9m$

To find the initial y-component of velocity we need to do the following steps:

$v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})$

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

$0=v_{oy}^{2}-2(9.8m/s^2)(2.9m)$

$v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s$

Since the bait is released at 25º

$v_{oy}=v_{o}sin(25)$

$v_{o}=\frac{7.53m/s}{sin(25)}=18m/s$

6 0

2 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.240 rev/s. The magnitude

bazaltina [42]

Explanation:

Given that,

Angular velocity = 0.240 rev/s

Angular acceleration = 0.917 rev/s²

Diameter = 0.720 m

(a). We need to calculate the angular velocity after time 0.203 s

Using equation of angular motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$\omega_{f}=0.240+0.917\times0.203$

$\omega_{f}=0.426\ rev/s$

The angular velocity is 0.426 rev/s.

(b). We need to calculate the tangential speed of the blade

Using formula of tangential speed

$v= r\omega$

Put the value into the formula

$v = \dfrac{0.720 }{2}\times0.426\times2\pi$

$v=0.963\ m/s$

The tangential speed of the blade is 0.963 m/s.

(c). We need to calculate the magnitude at of the tangential acceleration

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.36\times0.917\times2\pi$

$a_{c}=2.074\ m/s^2$

The tangential acceleration is 2.074 m/s².

Hence, This is required solution.

4 0

2 years ago

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