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3 years ago

1, Can bed bugs adapt to the heat treatments? Why or why not?

Physics

1 answer:

mart [117]3 years ago

6 0

Answer:

Heat radiates into walls and your belongings to get bed bugs in places that are hard to get to with insecticides Hope It Helps

Explanation:

Studies have shown that bed bugs can become resistant to chemicals and adapt to living in dangerous environments. This can make it a difficult job to treat for bed bugs with chemical products.

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You can see yourself in the mirror because?

jok3333 [9.3K]

Mirrors reflect light waves.

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3 years ago

Read 2 more answers

10. In Freudian theory, which one of the following is an example of a Thanatos instinct, which includes aggression, trauma, and

Lelechka [254]

D substance abuse :) hope this helps

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2 years ago

A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re

slamgirl [31]

Answer:

tension: 19.3 N
acceleration: 3.36 m/s^2

Explanation:

Given

mass A = 2.0 kg

mass B = 3.0 kg

θ = 40°

Find

The tension in the string

The acceleration of the masses

Solution

Mass A is being pulled down the inclined plane by a force due to gravity of ...

F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

F/m = a = F/m

(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

T = (2(29.4) +3(12.5986))/5 = 19.3192 N

Then the acceleration of B is ...

a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0

4 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$