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jarptica [38.1K]

3 years ago

10

Two vectors, X and Y, form a right angle. Vector X is 48 inches long and vector Y is 14 inches long. the length of the resultant

vector is____ inches.

1 answer:

Scorpion4ik [409]3 years ago

4 0

Solution is provided below hope this helps

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I have to write something here, so like hello and please help

ExtremeBDS [4]

Answer:

the extension would be less the new extension might be 3 cm

Explanation:

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2 years ago

Two technicians are discussing U-joints. Technician A says that a defective U-joint could cause a loud clunk when the transmissi

Nookie1986 [14]

Answer:

Technician A and Technician B are correct.

Explanation:

6 0

3 years ago

A 100-watt lightbulb uses 1 kilowatt-hour of electrical

Harlamova29_29 [7]

Answer:

Energy = 0.25 kilowatt-hour

Explanation:

Given the following data;

Power = 25 Watts

Time = 10 hours

Power can be defined as the energy required to do work per unit time.

Mathematically, it is given by the formula;

$Power = \frac {Energy}{time}$

To find the energy consumed;

Energy = power * time

Substituting into the formula, we have;

Energy = 25 * 10

Energy = 250 Watt-hour

To convert to kilowatt-hour, we would divide by 1000;

Energy = 250/1000

Energy = 0.25 kilowatt-hour

3 0

3 years ago

As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes thro

LiRa [457]

Answer:

$x = t$

$y = \frac{1}{3}t$

$z =t$

Explanation:

Given

$r(t) = f(t)i + g(t)j + h(t)k$ at $t = 0$

Point: $(f(t0), g(t0), h(t0))$

$r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk$ , $t0 = 1$ -- Missing Information

Required

Determine the parametric equations

$r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk$

Differentiate with respect to t

$r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k$

Let t = 1 (i.e $t0 = 1$ )

$r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k$

$r'(1) = i +\frac{3}{3^2}j + (0 + 1)k$

$r'(1) = i +\frac{3}{9}j + (1)k$

$r'(1) = i +\frac{1}{3}j + (1)k$

$r'(1) = i +\frac{1}{3}j + k$

To solve for x, y and z, we make use of:

$r(t) = f(t)i + g(t)j + h(t)k$

This implies that:

$r'(1)t = xi + yj + zk$

So, we have:

$xi + yj + zk = (i +\frac{1}{3}j + k)t$

$xi + yj + zk = it +\frac{1}{3}jt + kt$

By comparison:

$xi = it$

Divide by i

$x = t$

$yj = \frac{1}{3}jt$

Divide by j

$y = \frac{1}{3}t$

$zk = kt$

Divide by k

$z = t$

Hence, the parametric equations are:

$x = t$

$y = \frac{1}{3}t$

$z =t$

3 0

3 years ago

If a bullet loses 1/nth of its velocity while passing through a plank,then no of planks required to stop the bullet is?

Novay_Z [31]

Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing <span><span>1/nth</span><span>1/nth</span></span><span> of its velocity per plank (however, the fact that the question does not mention this assumption arguably makes the question ambiguous).

</span><span>With this assumption, the energy loss becomes
</span><span>
ΔE = <span>1/2 </span>m<span>v2 </span>− <span>1/2 </span>m <span><span>(<span>v−<span>v/n</span></span>) </span><span>2
</span></span></span>
and the number of planks <span>NN</span><span> becomes
</span>
N = <span><span><span>1/2</span>m<span>v2 /</span></span><span>ΔE </span></span>= <span><span>n2/ </span><span>2n−1
</span></span>
Otherwise, if you assume that the bullet loses <span><span>1/<span>nth</span></span><span>1/<span>nth</span></span></span><span> of its velocity per plank, then the answer is </span><span><span>N=∞</span></span><span><span>

</span>
</span>

8 0

3 years ago