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3 years ago

11

__________ is when a person becomes less sensitive to a drug. A. Tolerance B. Craving C. Addiction D. Dependence

1 answer:

Luden [163]3 years ago

4 0

A) Tolerance

Tolerance is developed after using a drug repeatedly, so the body adapts to it. Because of that, people who develop a tolerance would then need to use more of that drug to get the same effect.

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The answer is A and no matter how many times I tried I can't get it.

ryzh [129]

Imagine a skinny straw in the water, standing right over the hole. The WEIGHT of the water in that straw is the force on the tape. Now, the volume of water in the straw is (1 mm^2) times (20 cm). Once you have the volume, you can use the density and gravity to find the weight. And THAT's the force on the tape. If the tape can't hold that force, then it peels off and the water runs out through the hole. /// This is a pretty hard problem, because it involved mm^2, cm, and m^3. You have to be very very very careful with your units as you work through this one. If you've been struggling with it, I'm almost sure the problem is the units.

5 0

3 years ago

A ball thrown into the air has its greatest kinetic energy as it reaches the highest point in its path. please select the best a

postnew [5]

F - False.

Its greatest kinetic energy is at the point of release.

It has the least kinetic energy, zero, at its highest point in its path.

3 0

3 years ago

Read 2 more answers

A 10-cm-long spring is attached to theceiling. When a 2.0 kg mass is hung from it,the spring stretches to a length of 15 cm.a.Wh

alekssr [168]

(a) 392 N/m

Hook's law states that:

$F=k\Delta x$ (1)

where

F is the force exerted on the spring

k is the spring constant

$\Delta x$ is the stretching/compression of the spring

In this problem:

- The force exerted on the spring is equal to the weight of the block attached to the spring:

$F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N$

- The stretching of the spring is

$\Delta x=15 cm-10 cm=5 cm=0.05 m$

Solving eq.(1) for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{19.6 N}{0.05 m}=392 N/m$

(b) 17.5 cm

If a block of m = 3.0 kg is attached to the spring, the new force applied is

$F=mg=(3.0 kg)(9.8 m/s^2)=29.4 N$

And so, the stretch of the spring is

$\Delta x=\frac{F}{k}=\frac{29.4 N}{392 N/m}=0.075 m=7.5 cm$

And since the initial lenght of the spring is

$x_0 = 10 cm$

The final length will be

$x_f = x_0 +\Delta x=10 cm+7.5 cm=17.5 cm$

8 0

3 years ago

Read 2 more answers

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coeﬃcient

lara31 [8.8K]

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

6 0

3 years ago

Calculate the phase angle (in radians) for a circuit with a maximum voltage of 12 V and w-50 Hz. The voltage source is connected

Vinvika [58]

Answer:

The phase angle is 0.0180 rad.

(c) is correct option.

Explanation:

Given that,

Voltage = 12 V

Angular velocity = 50 Hz

Capacitance $C= 20\times10^{-2}\ F$

Inductance $L=20\times10^{-3}\ H$

Resistance $R= 50\ Omega$

We need to calculate the impedance

Using formula of impedance

$z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}$

$z=\sqrt{50^2+(50\times20\times10^{-3}-\dfrac{1}{50\times20\times10^{-2}})^2}$

$z=50.00$

We need to calculate the phase angle

Using formula of phase angle

$\theta=\cos^{-1}(\dfrac{R}{z})$

$\theta=\cos^{-1}(\dfrac{50}{50.00})$

$\theta=0.0180\ rad$

Hence, The phase angle is 0.0180 rad.

3 0

3 years ago