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3 years ago

12

Reviewing old tests and quizzes will not help you determine what will be on a test because teachers don’t like to give students

hints about what could be on future tests. Please select the best answer from the choices provided T or F

2 answers:

kifflom [539]3 years ago

5 0

Answer:

False

Explanation:

riadik2000 [5.3K]3 years ago

4 0

Answer:

It is false

Explanation:

You might be interested in

Which elements are metalloids

zhannawk [14.2K]

The six commonly recognised metalloids are boron, silicon, germanium, arsenic, antimony, and tellurium....

7 0

3 years ago

A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find

FrozenT [24]

Answer:

The time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

Explanation:

Given :

Current $I = 0.106$ A

Area of plate $A = 36 \times 10^{-4}$ $m^{2}$

Plate separation $d = 4 \times 10^{-3}$ m

(A)

First find the capacitance of capacitor,

$C = \frac{\epsilon _{o} A }{d}$

Where $\epsilon _{o} = 8.85 \times 10^{-12}$

$C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4} }{4 \times 10^{-3} }$

$C = 7.9 \times 10^{-12}$ F

But $C = \frac{Q}{V}$

Where $Q = It$

$C = \frac{It}{V}$

$V = \frac{It}{C}$

Now differentiate above equation wrt. time,

$\frac{dV}{dt} = \frac{I}{C}$

$= \frac{0.106}{7.9 \times 10^{-12} }$

$= 1.34 \times 10^{10}$ $\frac{V}{s}$

Therefore, the time rate of change of flux is $1.34 \times 10^{10}$ $\frac{V}{s}$

8 0

3 years ago

8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h

pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0

3 years ago

An example of a folkway violation would be driving 70 mph in a 50-mph zone. Group of answer choices True False

shutvik [7]

Answer:

Yes.

Explanation:

That is also a law violation.

3 0

3 years ago

A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th

algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

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

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 $m/s^{2}$

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

$0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters$

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

$v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s$

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

$v=u+at\\\\t=\frac{v-u}{a}\\\\t=\frac{10.28-20}{-9.81}=0.991seconds\approx 1second$

4 0

3 years ago