Ask question

Ask question

All categories

3 years ago

9

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

0 m from rest. (You can check your answer by solving this problem using Newton's laws.)

1 answer:

fiasKO [112]3 years ago

8 0

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

You might be interested in

In a 350-m race, runner A starts from rest and accelerates at 1.6 m/s^2 for the first 30 m and then runs at constant speed. Runn

kifflom [539]

Answer:

B can take 0.64 sec for the longest nap .

Explanation:

Given that,

Total distance = 350 m

Acceleration of A = 1.6 m/s²

Distance = 30 m

Acceleration of B = 2.0 m/s²

We need to calculate the time for A

Using equation of motion

$s=ut+\dfrac{1}{2}at_{A}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times1.6\times t_{A}^2$

$t_{A}=\sqrt{\dfrac{30\times2}{1.6}}$

$t_{A}=6.12\ sec$

We need to calculate the time for B

Using equation of motion

$s=ut+\dfrac{1}{2}at_{B}^2$

Put the value in the equation

$30=0+\dfrac{1}{2}\times2.0\times t_{B}^2$

$t_{B}=\sqrt{\dfrac{30\times2}{2.0}}$

$t_{B}=5.48\ sec$

We need to calculate the time for longest nap

Using formula for difference of time

$t'=t_{A}-t_{B}$

$t'=6.12-5.48$

$t'=0.64\ s$

Hence, B can take 0.64 sec for the longest nap .

4 0

3 years ago

A motorbike reaches a speed of 20 m/s over 60m, whilst

Fynjy0 [20]

Initial speed = 2√10 m/s

<h3>Further explanation </h3>

Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration

An equation of uniformly accelerated motion

V = vo + at

Vt² = vo² + 2a (x-xo)

x = distance on t

vo / vi = initial speed

vt / vf = speed on t / final speed

a = acceleration

vf=20 m/s

d = 60 m

a = 3 m/s²

$\tt vf^2=vi^2+2.ad\\\\20^2=vi^2+2\times 3\times 60\\\\400=vi^2+360\\\\40=vi^2\\\\vi=\sqrt{40}=2\sqrt{10}~m/s$

7 0

3 years ago

A wave has an amplitude of 0.2 cm.

Anettt [7]

The distance between the resting point and maximum height of the wave is 0.2 cm.

The amplitude is measured from the resting point up to the highest point of the wave.

3 0

2 years ago

The uniform 100-kg I-beam is supported initially by its end rollers on the horizontal surface at A and B. by means of the cable

Ann [662]

I attached a free body diagram for a better understanding of this problem.

We start making summation of Moments in A,

$\sum M_A = 0$

$P(6cos\theta)-981(4cos\theta)=0$

$P=654N$

Then we make a summation of Forces in Y,

$\sum F_y = 0$

$654+R-981 = 0$

$R=327N$

At the end we calculate the angle with the sin.

$sin\theta = \frac{3m}{4m+2m+2m} = \frac{3m}{8m}$

$\theta = 22.02\°$

8 0

3 years ago

Please help ASAP A uniform disk of mass M and radius R is pivoted such that it can rotate freely about a horizontal axis through

kompoz [17]

Answer: F = mg(1 + 4m / (½M + m))

Explanation:

"At this point seems" unclear. If the particle is at the top of the disc and angular velocity is negligible, then the force would equal the weight of the particle. F = mg

The more interesting question would be what force is needed to keep the particle attached when significant angular rotation has been achieved. The maximum point would be diametrically opposed to the starting point.

I will analyze it there

The potential energy will convert to kinetic energy

mgh = ½Iω²

mg(2R) = ½(½MR² + mR²)ω²

4mgR = R²(½M + m)ω²

ω² = 4mg / (R(½M + m))

With m at the lowest position, the force of attachment must support the weight of m and provide for the needed centripetal acceleration

F = m(g + ω²R)

F = m(g + 4mg / (R(½M + m))R)

F = mg(1 + 4m / (½M + m))

7 0

2 years ago