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3 years ago

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Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the

finish line with a final velocity v. The losing car accelerates at a constant rate a/4. How far had the losing car traveled, d, when the winning car crossed the finish line?

1 answer:

Mila [183]3 years ago

5 0

Answer: $d=\frac{v^2}{8a}$

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

$v=at$

$t=\frac{v}{a}$

Now loosing car is accelerating with $\frac{a}{4}$

Distance traveled by loosing car in time t

$s_1=ut+\frac{at^2}{2\cdot 4}$

$s_1=0+\frac{a}{8}\times (\frac{v}{a})^2$

$s_1=\frac{v^2}{8a}$

Thus distance d traveled by loosing car is given by $d=\frac{v^2}{8a}$

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A uniform brick of length 21 m is placed over

Paha777 [63]

Answer:

15.75 m

Explanation:

First, let's look at the top brick by itself. In order for it not to tip over the bottom brick, its center of gravity must be right at the edge of the bottom brick. So the edge of the top brick must be 10.5 m from the edge of the bottom brick.

Now let's look at both bricks as a combined mass. We know the total length of this combined brick is 10.5 m + 21 m = 31.5 m. And we know that for it to not tip over the edge of the surface, its center of gravity must be at the edge. So the edge of the combined brick must be 31.5 m / 2 = 15.75 m from the edge of the surface.

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3 years ago

Importance of simple machines pleasecgive answer in points

oksian1 [2.3K]

Answer:

Explanation:

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2 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

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3 years ago

A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of

Sauron [17]

Answer:

The value is $t = 689.029 \ hours$

Explanation:

From the question we are told that

The molar mass of hydrazine is $Z = 32 g/mol = \frac{32}{1000} = 0.032 \ kg/mol$

The initial temperature is $T_i = -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K$

The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

The specific heat capacity is $c_h = 0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)$

The power available is $P = 300 \ W$

The mass of the fuel is $m = 1640 \ kg$

Generally the number of moles of hydrazine present is

$n = \frac{m}{Z}$

=> $n = \frac{1640}{= 0.032}$

=> $n = 51250 \ mol$

Generally the quantity of heat energy needed is mathematically represented as

$Q = n * c_h * (T_f -T_i)$

=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

=> $Q = 7.441516913 * 10^{8} \ J$

Generally the time taken is mathematically represented as

$t = \frac{Q}{P}$

=> $t = \frac{7.441516913 * 10^{8} }{300}$

=> t = 2480505.6377 s

Converting to hours

$t = \frac{2480505.6377}{3600}$

=> $t = 689.029 \ hours$

6 0

2 years ago

A concave mirror always forms a virtual image of a real object. O True O False

kifflom [539]

Answer:

The given statement is false.

Explanation:

The spherical mirrors are the mirror that are a part of a sphere. Concave and convex mirrors are two types of spherical mirrors.

A concave mirror always forms real and inverted image. A convex mirror forms real and virtual images.

For concave mirror, the value of magnification is less that 1. Also, the focal length is negative for concave mirrors.

So, the given statement is false as a concave mirror always forms a real and inverted image. Hence, this is the required solution.

4 0

3 years ago