A lab cart is loaded with different masses and moved at various velocities

Greg throws a 2.8-kg pumpkin horizontally off the top of the school roof in order to hit Mr. H's car. The car has parked a dista

Igoryamba

Answer:

The horizontal velocity is $v = 9.2 m/s$

Explanation:

From the question we are told that

The mass of the pumpkin is $m = 2.8 \ kg$

The distance of the the car from the building's base is $d = 13.4 \ m$

The height of the roof is $h = 10.4 \ m$

The height is mathematically represented as

$h = \frac{1}{2} gt^2$

Where g is the acceleration due to gravity which has a value of $g =9.8 \ m/s^2$

substituting values

$10.4= 0.5 * 9.8 * t$

making the time taken the subject of the formula

$t = \frac{10.4}{0.5 * 9.8 }$

$t = 1.457 \ s$

The speed at which the pumpkin move horizontally can be represented mathematically as

$v = \frac{d}{t}$

substituting values

$v =\frac{13.4}{1.457}$

$v = 9.2 m/s$

7 0

3 years ago

A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result

vivado [14]

Answer:

$21.6\ \text{kg m}^2$

$3.672\ \text{Nm}$

$54.66\ \text{revolutions}$

Explanation:

$\tau$ = Torque = 36.5 Nm

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2$

Torque is given by

$\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2$

The wheel's moment of inertia is $21.6\ \text{kg m}^2$

t = 60.6 s

$\omega_i$ = 10.3 rad/s

$\omega_f$ = 0

$\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2$

Frictional torque is given by

$\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}$

The magnitude of the torque caused by friction is $3.672\ \text{Nm}$

Speeding up

$\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}$

Slowing down

$\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}$

Total number of revolutions

$\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}$

$\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}$

The total number of revolutions the wheel goes through is $54.66\ \text{revolutions}$ .

3 0

3 years ago

Hartman value profile

Grace [21]

What is your question exactly? I'm confused?

6 0

3 years ago

How long does it take the moon to go through an entire set of phases?.

Bas_tet [7]

It take approximately 29.5 days for moon to do its entire set of phases.

<h3>Explanation</h3>

The Moon is the only natural satellite of the Earth which undergoes three motions, that is :

Rotating on its own axis
Evolving around the Earth
Together with the Earth evolving around the sun as the center of the solar system

With that, the moon has two periods of revolution, namely:

Sidereal revolution, which is the original revolution of the Moon. This sidereal revolution is really the time it takes the Moon to orbit the Earth. The sidereal revolution of the moon has a time span of 27.3 days or more accurate is approximately 27 days, 7.72 hours.
Synodic revolution, namely the revolution of the Moon as seen from Earth as a series of moon phases (from the new moon phase, to the next new moon phase). The synodic revolution is slower, because the Moon needs to catch up with the Earth rotating in the same direction as the Moon. The synodic revolution of the moon has a time span of 29.5 days or to be more accurate approx 29 days, 12.734 hours.

6 0

2 years ago

A ball is thrown upward with a speed of 28.2 m/s.A. What is its maximum height?B. How long is the ball in the air?C. When does t

Ede4ka [16]

Answer:

(A) The maximum height of the ball is 40.57 m

(B) Time spent by the ball on air is 5.76 s

(C) at 33.23 m the speed will be 12 m/s

Explanation:

Given;

initial velocity of the ball, u = 28.2 m/s

(A) The maximum height

At maximum height, the final velocity, v = 0

v² = u² -2gh

u² = 2gh

$h = \frac{u^2}{2g}\\\\h = \frac{(28.2)^2}{2*9.8}\\\\h = 40.57 \ m$

(B) Time spent by the ball on air

Time of flight = Time to reach maximum height + time to hit ground.

Time to reach maximum height = time to hit ground.

Time to reach maximum height is given by;

v = u - gt

u = gt

$t = \frac{u}{g}$

Time of flight, T = 2t

$T = \frac{2u}{g}\\\\ T = \frac{2*28.2}{9.8}\\\\ T = 5.76 \ s$

(C) the position of the ball at 12 m/s

As the ball moves upwards, the speed drops, then the height of the ball when the speed drops to 12m/s will be calculated by applying the equation below.

v² = u² - 2gh

12² = 28.2² - 2(9.8)h

12² - 28.2² = - 2(9.8)h

-651.24 = -19.6h

h = 651.24 / 19.6

h = 33.23 m

Thus, at 33.23 m the speed will be 12 m/s

6 0

3 years ago