What should #1 be labeled?

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling

solniwko [45]

Answer:

17.86 m

Explanation:

The first step is to find the velocity of the swimmer.

Velocity, v = s/t

Velocity, v = 5 / 0.27

Velocity, v = 18.52 m/s

Then, we find the height

h = ½gt²

h = ½ * 9.8 * 0.27²

h = 4.9 * 0.0729

h = 0.36 m

Next, we apply the law of conservation of energy

mgH = mgh + ½ mv²

9.8Hm = m(9.8 * 0.36 + 0.5 * 18.52²)

9.8H = 3.528 + 171.4952

9.8H = 175.02

H = 175.02 / 9.8

H = 17.86 m

Therefore, the needed height is 17.86 m

8 0

4 years ago

7.

laiz [17]

7. Matthias Schleiden

8. An Element

9. Compound Microscope

6 0

3 years ago

A thread holds two carts together on a frictionless surface. A spring is compressed

Goryan [66]

Answer:

Velocity on the right side of the cart $=0.09\ ms^{-1}$

Explanation:

Given

⇒The mass on the left of the cart $m_1=1.5\ kg$

Its velocity $v_1=27\ cm/s$ , $v_1=\frac{27}{100}=0.27\ m/s$

⇒Mass on the right of the cart $m_2=4.5\ kg$

Velocity $=?$ We have to find $v_2$

From

The law of conservation of linear momentum:

We can say that.

Initial momentum will equalize the final momentum.

And momentum is the product of mass and its velocity.

Assigning one of its velocity as negative because both are in different direction.

Lets call $v_1=-0.27m/s$

Recalling the formula and plugging the values.

$m_1(-v_1)+m_2v_2=0$

$v_2=-\frac{m_1(-v_1)}{m_2} =-\frac{1.5\times -0.27}{4.5} =0.09\ m/s$

So the velocity of the cart on the right side that has a mass of $4.5\ kg$ is $0.09\ ms^{-1}$

4 0

4 years ago

In some cases fixture wires may be used for

zalisa [80]

You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.

<h3>What are some uses of fixture wires?</h3>

Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG

In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

To know more about fixture wires, refer

brainly.com/question/26098282

#SPJ4

3 0

1 year 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