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2 years ago

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You attach a block to the bottom end of a spring hanging vertically. You slowly let the block move down and find that it hangs a

t rest with the spring stretched by 18.0 cm. Next, you lift the block back up and release it from rest with the spring unstretched. What maximum distance does it move down

1 answer:

yKpoI14uk [10]2 years ago

5 0

Answer:36 cm

Explanation:

Given

At Equilibrium extension a is 18 cm

suppose m is the mass of block

thus

$ka=mg$

$k=\frac{mg}{a}$

$k=\frac{mg}{0.18}$

Now if the block is released from un-stretched position

conserving Energy and supposing block moves x units down

$\frac{1}{2}kx^2=mgx$

$x=\frac{2mg}{k}$

$x=2mg\times \frac{0.18}{mg}$

$x=0.36 \approx 36 cm$

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A 64 kg swimmer jumps, with a velocity of 4.2 m/s, off the front of a 25 kg kayak when the kayak is moving forward at a velocity

Crank

Answer:

3.88m/s

Explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and 2 are the initial velocities

v is the final velocity

Given

m1 = 64kg

u1 = 4.2m/s

m2 = 25kg

u2 = 3.2m/s

Required

Final velocity v

Substitute the given values into the formula

64(4.2)+25(3.2) = (65+25)v

268.8+80 = 90v

348.8 = 90v

v = 348.8/90

v = 3.88m/s

Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s

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2 years ago

(Fill the blank)<br> the ________ g____ makes saliva that make your food wet and easy to swallow

blagie [28]

Answer:

What make saliva that make your food wet and easy to swallow?

enzyme amylase

Explanation:

The digestive functions of saliva include moistening food, and helping to create a food bolus, so it can be swallowed easily. Saliva contains the enzyme amylase that breaks some starches down into maltose and dextrin. Thus, digestion of food occurs within the mouth, even before food reaches the stomach.

<em>Hope </em><em>it </em><em>h</em><em>elp </em><em>4</em><em>u</em>

<em>Y</em><em>an </em><em>na </em><em>baby</em>

6 0

2 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

2 years ago

Many Amtrak trains can travel at a top speed of 42.0 m/s. Assuming a train maintains that speed for several hours, how many kilo

777dan777 [17]

Answer:

605 km

Explanation:

Hello

the same units of measure should be used, then

Step 1

convert 42 m/s ⇒ km/h

1 km =1000 m

1 h = 36000 sec

$42 \frac{m}{s}*\frac{1\ km}{1000\ m}=0.042\ \frac{km}{s}\\ 0.042\ \frac{km}{s}\\$

$0.042\ \frac{km}{s}*\frac{3600\ s}{1\ h} =151.2 \frac{km}{h}\\ \\Velocity =151.2\ \frac{km}{h}$

Step 2

find kilometers traveled after 4 hours

$V=\frac{s}{t}\\ \\$

V,velocity

s, distance traveled

t. time

now, isolating s

$V=\frac{s}{t} \\s=V * t\\$

and replacing

$s=V * t\\s=151.2\frac{km}{h}*4 hours\\ s=604.8 km\\$

S=604.8 Km

Have a great day

4 0

2 years ago

A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago