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3 years ago

Which of these is a key feature of an experimental study?

2 answers:

7 0

The key feature in the experimental study is C. <span>The treatment in the experiment must be applied to each of the individuals in the experimental group. This is because it is made sure that the variables and conditions in different correspondents are applied so that actual results may be concluded.</span>

juin [17]3 years ago

3 0

the only logical answer is C

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A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spe

saw5 [17]

Answer:

Mass of ion will be $22\times 10^{-13}kg$

Explanation:

We have given ion is triply charged that is $q=3\times 1.6\times 10^{-19}=4.8\times 10^{-19}C$

Radius r = 36 cm = 0.36 m

Velocity of the electron $v=7\times 10^6m/sec$

Magnetic field B = 0.55 T

We know that radius of the path is given by $r=\frac{mv}{qB}$

$m=\frac{rqB}{v}=\frac{0.36\times 4.8\times 10^{-19}\times 7\times 10^6}{0.55}=22\times 10^{-13}kg$

6 0

3 years ago

An iron ball is dropped at a height of 10 m from the surface of the moon.

galben [10]

Answer:

3.51s

Explanation:

There are many students who can not get answers step by step and on time

So there are a wats up group where you can get help step by step and well explained by the trusted experts.

3 0

3 years ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

A swimmer jumps from a waterfall thats is 8.5 m high from the lake below. How fast must he run horizontally to dive 2.5m away fr

Mashutka [201]

He must run at 6m/s

6 0

3 years ago

A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c

natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

$mg = \rho V_{displaced} g$

$V_{displaced} = \frac{m}{\rho}$

$V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3$

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

$V\rho_w = m_{ice}$

$V = \frac{0.200}{1000} = 2\times 10^{-4} m^3$

So here extra volume that rise in the level will be given as

$\Dleta V = V - V_{displaced}$

$\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}$

$(\pi (0.039^2) h = 0.41 \times 10^{-4}$

$h = 0.86 cm$

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0

3 years ago