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2 years ago

15

Guys pls thi is my last pointa just answer this!!!!(WILL GIVE BRAINLY) A student takes apart a wooden box. What can they build w

ith the materials from the box?
A) They cannot make anything with the materials.
B) They can make something new with the materials.
C) They can only make a wooden box from the materials.

1 answer:

bazaltina [42]2 years ago

5 0

Answer:

i guess u could pick "B" :)

Explanation:

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A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

8 0

3 years ago

Friction provides the force needed for a car to travel around a flat, circular race track. What is the maximum speed at which a

blagie [28]

Answer:

The maximum speed at which the car can safety travel around the track is 18.6m/s.

Explanation:

Since the car is in circular motion, there has to be a centripetal force $F_c$ . In this case, the only force that applies for that is the static frictional force $f_s$ between the tires and the track. Then, we can write that:

$f_s=F_c$

And since $f_s\leq \mu N$ and $F_c=\frac{mv^{2}}{r}$ , we have:

$\mu N\geq \frac{mv^{2}}{r}$

Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:

$N-mg=0\\\\\\implies N=mg$

Substituting this expression for $N$ and solving for $v$ , we get:

$\mu mg\geq \frac{mv^{2}}{r}\\\\v\leq \sqrt{\mu gr}$

Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:

$v\leq \sqrt{(0.42)(9.81m/s^{2})(84.0m)}\\\\v\leq 18.6m/s$

It means that in its maximum value, the speed of the car is equal to 18.6m/s.

7 0

2 years ago

Read 2 more answers

If a one-pound weight were one foot from the pivot or shaft of a lever, how many pound-feet of force would result? A. 4 B. 1 C.

mixas84 [53]

The force is still 1 pound.

The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)

8 0

2 years ago

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What are the features of image formed by convex mirror.

STatiana [176]

The image is always virtual and erect. The image is highly diminished or point sized. It is always formed between F and P.

4 0

3 years ago

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What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

7 0

3 years ago