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3 years ago

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Question 7 of 10 A strong magnet is used to separate certain items from other recyclables Which two items found in recyclables w

ould be attracted to the magnet? O A. Cardboard boxes o B. Iron cooking pans O c. Glass mixing bowls O D. Flexible refrigerator magnets

1 answer:

timama [110]3 years ago

6 0

Answer:

D. Flexible refrigerator magnets

B. Iron cooking pans

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Silk flexibility would be considered a trait. TRUE OR FALSE

wel

The answer would be true.

Explanation:
Although some silk in is more flexible than others, there are many varieties. There are high medium and low flexibility. Even though there are different levels, it is still considered a trait. Again, the answer in True.

Hope this helped you. I did some research on this before, so this should be correct. If you aren’t satisfied then you can ask me a question! Have a nice day.
:)

6 0

3 years ago

Read 2 more answers

Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B

pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0

3 years ago

A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and

N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0

4 years ago

Read 2 more answers

An ideal gas, initially at a pressure of 11.2 atm and a temperature of 299 K, is allowed to expand adiabatically until its volum

Tju [1.3M]

Answer:

The pressure is $P_2 = 4.25 \ a.t.m$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 11.2\ a.t.m$

The temperature is $T_1 = 299 \ K$

Let the first volume be $V_1$ Then the final volume will be $2 V_1$

Generally for a diatomic gas

$P_1 V_1 ^r = P_2 V_2 ^r$

Here r is the radius of the molecules which is mathematically represented as

$r = \frac{C_p}{C_v}$

Where $C_p \ and\ C_v$ are the molar specific heat of a gas at constant pressure and the molar specific heat of a gas at constant volume with values

$C_p=7 \ and\ C_v=5$

=> $r = \frac{7}{5}$

=> $11.2*( V_1 ^{\frac{7}{5} } ) = P_2 * (2 V_1 ^{\frac{7}{5} } )$

=> $P_2 = [\frac{1}{2} ]^{\frac{7}{5} } * 11.2$

=> $P_2 = 4.25 \ a.t.m$

8 0

3 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of <u>2.1 m</u>

7 0

3 years ago