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13 days ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Physics

2 answers:

inna [987]13 days ago

7 0

The gradient of a line is a numerical value that indicates both the angle and incline of the line. On a distance versus time graph, this value represents the speed.

At t1=2.0 seconds
v = 7/3 meters per second

At t2=4.0 seconds
v = 0 meters per second

At t3=13 seconds
v = -0.6 meters per second

I hope this clarifies your question. Have a wonderful day.

Softa [913]13 days ago

7 0

As this graph shows distance versus time, the object's speed at any moment corresponds to the slope of the graph's section directly above that instant on the x-axis.

At t1 = 2.0 s
This point lies halfway through the initial portion of the graph from 0 to 3 seconds.
The slope here equals 7 divided by 3. v1 = 7/3 m/s.

At t2 = 4.0 s
This moment is located at the center of the flat segment spanning 3 to 6 seconds.
The slope during this interval is 0. v2 = 0.

At t3 = 13 s
This time falls in the middle of the descending segment from 11 to 16 seconds.
The slope for this segment is -3 divided by 5. v3 = -0.6 m/s.

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The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring

Sav [1095]

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

$E=PE_{2}-PE_{1}$

$E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2$

Substitute the values into the formula

$E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2$

$E=0.0072\ J$

$E=7.2\ mJ$

Therefore, a total of 7.2 mJ is needed to extend the pen.

7 0

11 days ago

Read 2 more answers

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Sav [1095]

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

7 0

3 days ago

The amount of electric energy consumed by a 60.0-watt lightbulb for 1.00 minute could lift a

Sav [1095]

Answer:

Explanation:

For a 60W light bulb used for 1 minute:

P = 60 W

t = 1 minute = 60 seconds

This energy is capable of lifting an object weighing 10N.

W = 10N

This indicates conversion of electrical energy into potential energy.

Let's calculate the electrical energy:

Power describes the rate of work done.

Power = Work / time

Thus, work = power × time

Work = 60 × 60

Work = 3600 J

Potential energy calculation:

P.E = mgh

Where the weight is given by:

W = mg

Therefore, P.E = W·h

P.E = 10·h

Thus, we equate:

Potential energy = Electrical energy

P.E = Work

10·h = 3600

Dividing both sides by 10 gives:

h = 3600 / 10

h = 360m

The object can be lifted to a height of 360m.

6 0

6 days ago

A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it

Ostrovityanka [942]

Answer:

0.130

Explanation:

The coefficients of static friction recorded for each trial are listed as follows:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

Adding these coefficients together results in: 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

= 0.779

Consequently;

the mean coefficient of static friction = $\frac{sum of coefficient of static friction}{number of trials}$

= $\frac{0.779}{6}$

= 0.12983

The mean coefficient of static friction is 0.130

8 0

6 days ago

A transformer is to be designed to increase the 30 kV-rms output of a generator to the transmission-line voltage of 345 kV-rms.

inna [987]

Answer:

$n_s = 920 \turns$

Explanation:

Given,

Voltage of the primary coil (V_p) = 30 kV-rms

Voltage of the secondary coils (V_s) = 345 kV-rms

number of turns in the primary coil (n_p) = 80 turns

number of turns in the secondary coil (n_s) =?

the ratio of turns between primary and secondary coils

$\dfrac{n_p}{n_s} = \dfrac{V_p}{V_s}$

$\dfrac{n_s}{n_p} = \dfrac{V_s}{V_p}$

$n_s = n_p \dfrac{V_s}{V_p}$

$n_s = 80\times \dfrac{345}{30}$

$n_s = 80\times 11.5$

$n_s = 920 \turns$

The number of turns in the secondary coil is equal to $n_s = 920 \turns$

4 0

1 day ago