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3 years ago

11

The speed of a car decreases uniformly as it passes a curve point where normal component of acceleration is 4 ft/sec2. If the ca

r total acceleration of 5ft/sec2 is the same as it passes a hump, the tangential component of acceleration is _______________ ft/sec2.

1 answer:

belka [17]3 years ago

8 0

Answer:

45

Explanation:

ft/sec2

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A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

3 years ago

Take another look at lines 2 and 3. Suppose you use distance and time between any pair of neighboring dots to calculate speed:

Nataly [62]

Answer:

Add the two speeds together.

Then, divide the sum by two. This will give you the average speed for the entire trip. So, if Ben traveled 40 mph for 2 hours, then 60 mph for another 2 hours, his average speed is 50 mph.

8 0

3 years ago

Read 2 more answers

Pleaseeee Please help, I will love you forever and ever

matrenka [14]

Answer:

The answer to your question is

Explanation:

Data

mass = 0.5kg

T1 = 35

T2 = ?

Q = - 6.3 x 10⁴ J = - 63000 J

Cp = 4184 J / kg°C

Formula

Q = mCp(T2 - T1)

T2 = T1 + Q/mCp

Substitution

T2 = 35 - 63000/(0.5 x 4184)

T2 = 35 - 63000/2092

T2 = 35 - 30.1

T2 = 4.9 °C

6 0

3 years ago

In preparation for an electrophoresis procedure, enzymes are added to DNA in order to

lapo4ka [179]

It is 2.) Cut the DNA into fragments

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3 years ago

A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc

Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity $v_1$ in time $t_1$

$\frac{d}{2}=v_1t_1----1$

from B to C

distance traveled is 0.5 d with $v_2$ and $v_3$ velocity for half-half time

$\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2$

divide 1 and 2 we get

$\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}$

$\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}$

Now average velocity is given by

$v_{avg}=\frac{d}{t_1+t_2}$

taking $t_1$ common

$v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}$

$v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}$

$v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}$

6 0

3 years ago