9. If 100 J of work is done in 20 second the power is ..

The stopwatch used by a student to measure velocity of a pulse in a slinky was of least count 0.1 second. He stops the stopwatch

sweet-ann [11.9K]

Least count of the pulse stopwatch is given by

$\Delta t = 0.1 s$

this means each division of the stopwatch will measure 0.1 s of time

After 3 journeys from one end to other we can see that total time that is measured here is shown by the clock as 52nd division

So here total time is given as

Time = (Number of division) (Least count)

now we will have

$T = 52 \times 0.1s$

$T = 5.2 s$

4 0

3 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago

transmission electron microscopes that use high-energy electrons accelerated over a range from 40.0 to 100 kv are employed in ma

Gekata [30.6K]

The spatial limitations in Picometer for the given range of electrons would be around 50 picometers.

What is a transmission electron microscope?

A transmission electron microscope (TEM) is a type of microscope that uses a beam of high-energy electrons to produce detailed images of the structure of materials at the atomic or molecular scale. TEMs work by passing a focused beam of electrons through a thin sample and collecting the transmitted electrons on a fluorescent screen or an electronic detector. The interaction of the sample with the electrons results in the formation of an image that can be magnified and displayed on a computer monitor. TEMs are widely used in the fields of materials science, biology, and nanotechnology and can provide information about the structure, composition, and properties of materials with a high level and resolution.

According to the problem:

The spatial resolution of a transmission electron microscope (TEM) is determined by the size of the electron probe, which is directly related to the energy of the electrons. The higher the energy of the electrons is, the smaller the size of the probe is and the higher the spatial resolution.

At the lower end of the energy range of 40.0 kV, the spatial resolution of the TEM would be on the order of hundreds of nanometers. At the higher end of the range (100 kV), the spatial resolution would be on the order of tens of nanometers.

In general, TEMs with electron energy in the range of 40-100 kV are capable of resolving details down to around 50 picometers (pm). However, the actual spatial resolution will depend on various factors, such as the quality of the electron optics, the stability of the electron beam, and the sample preparation.

It's worth noting that TEMs with even higher electron energies (up to several hundred kV) are available, which can achieve spatial resolutions down to the sub-angstrom level (less than 0.1 pm). However, these instruments are much more expensive and complex to operate than TEMs with lower electron energies.

To know more about de broglie wavelength, visit:

brainly.com/question/17295250

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7 0

7 months ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

$F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N$

The force applied by the left support is 2123.33 N

8 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

3 years ago