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gavmur [86]
2 years ago
7

How much force is required to pull a spring 3.0 cm from

Physics
1 answer:
avanturin [10]2 years ago
4 0

Answer:

I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.

I hope it helps

plz let me know if it is wrong or right.

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f a car is speeding down a road at 40 miles/hour (mph), how long is the stopping distance D40 compared to the stopping distance
Oksana_A [137]

Answer:

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

Explanation:

given data

speed = 40 miles / hour

distance = D40

speed limit = 25 miles / hour

distance = D25

to find out

express number a multiple of stopping distance @ 25 mph

solution

we know here stopping distance is directly proportional to (speed)²

so here speed ratio is

initial speed = \frac{40}{25}

so initial speed = 1.6

so

stopping distance increase = (1.6)²

\frac{D40}{D25} = (1.6)²

\frac{D40}{D25} = 2.56

so here

D40 = 2.56 × D25

so number is 2.56 multiple of stopping distance @ 25 mph

5 0
1 year ago
Atoms can be "cooled" to incredibly low temperatures by letting them interact with a laser beam. Various novel quantum phenomena
Oksanka [162]

Answer:

the rms speed of cesium atoms that have been cooled to a temperature of 100nK = 0.43cm/s or 0.0043m/s

Explanation:

The concept of root mean square velocity is applied, where the average translational kinetic is related to the actual kinetic energy, the expression for the root mean square is the generated.

The detailed steps and appropriate substitution is as shown in the attachment.

8 0
2 years ago
A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha
Gelneren [198K]

Answer:

a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

Explanation:

Given data

velocity v₀=20 cm/s at time t=3s

velocity vf=0 at time t=8 s

To find

Average Acceleration at time=3s to 8s

Solution

As we know that acceleration is first derivative of velocity with respect to time

a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\  a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\  a_{acceleriation}=-4cm/s^{2}\\ or\\  a_{acceleriation}=-0.04m/s^{2}

8 0
2 years ago
You will now use Graphical Analysis to calculate the slope of the line. Choose Analyze from the menu. Select Automatic Curve Fit
Vlad1618 [11]
The value for the slope is <span>M=1.13</span>
3 0
1 year ago
A baseball is hit with a speed of 33.6 m/s. Calculate its height and the distance traveled if it is hit at angles of 30.0°, 45.0
Solnce55 [7]

Answer:

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

Explanation:

Let ∝ be the angle with respect to the horizontal that the baseball is hit with.

The horizontal component of the velocity is vcos(∝) and the vertical component of the velocity is vsin(∝).

Ignore air resistant, only gravitational acceleration g = -9.81 m/s2 affect the ball vertically. We can use the following equation to calculate the time it takes to reach maximum height (at 0 speed)

vsin(\alpha) + gt = 0

t = \frac{-vsin(\alpha)}{g}

So the vertical distance it travels within time t is

y = vsin(\alpha)t + gt^2/2 = vsin(\alpha)\frac{-vsin(\alpha)}{g} + g\frac{(-vsin(\alpha))^2}{2g^2}

y = \frac{-v^2sin^2(\alpha)}{g} + frac{v^2sin^2(\alpha)}{2g}

y = \frac{-v^2sin^2(\alpha)}{2g}

Similarly the horizontal distance it travels within time t is:

x = vcos(\alpha)t = vcos(\alpha)\frac{-vsin(\alpha)}{g}

x = \frac{-v^2sin(2\alpha)}{2g}

We can pre-calcualte the quantity \frac{-v^2}{2g} = \frac{-33.6^2}{2*(-9.81)} = 57.54

So y = 57.54sin^2(\alpha)

x = 57.54sin(2\alpha)

From here we can plug-in the angles values of 30, 45 and 60 degrees

At 30 degrees, height = 14.39 m and distance = 49.83 m

At 45 degrees, height = 28.77 m and distance = 57.54 m

At 60 degrees, height = 43.16 m and distance = 49.83 m

8 0
1 year ago
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