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3 years ago

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An athlete whirls a 8.00 kg hammer m from the axis of rotation in a horizontal circle, as shown in the figure below . The hammer

makes one revolution in s. What is the centripetal force on the hammer ?

1 answer:

Zigmanuir [339]3 years ago

4 0

Answer:

$mv \div r \\ mass \times velocity \div radius$

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In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.

alisha [4.7K]

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case $\Delta U=250kJ$ , $\Delta Q$ is the heat tranferred, and $W$ is the work, $W=-480kJ$ with a negative sign since the work is done by the system.

From the previous equation we solve for heat, because it is the unknown variable in this problem

$\Delta Q=\Delta U +W$

And replacing the known values:

$\Delta Q=250kJ +(-480kJ)$

$\Delta Q=250kJ -480kJ$

$\Delta Q=-230kJ$

The negative sign shows us that the heat is tranferred from the system into the surroundings.

3 0

3 years ago

Two converging lenses are placed 30 cm apart. The focal length of the lens on the right is 20 cm while the focal length of the l

Masja [62]

Answer:

a) I2 = 3 (o-10) / (o- 30) , b) h ’/h= 3 (o-10) / o (o-30)

Explanation:

The builder's equation is

1 / f = 1 / o + 1 / i

Where f is the focal length, or e i are the distance to the object and image, respectively

As the separation between the lenses is greater than the focal distances, we must work them individually and separately. Let's start with the leftmost lens with focal length f = 15 cm

Let's calculate the position of the image of this lens

1 / i1 = 1 / f - 1 / o

1 / i1 = 1/15 - 1 / o

i1 = o 15 / (o-15)

Let's calculate the distance to the image of the second lens, for this the image of the first is the distance to the object of the second

o2 = d-i1

We write the builder equation

1 / f2 = 1 / o2 + 1 / i2

1 / i2 = 1 / f2 -1 / o2

1 / i2 = 1 / f2 - 1 / (d-i1)

1 / i2 = 1/20 - 1 / (d-i1) (1)

Let's evaluate the last term

d-i1 = d - 15 o / (o-15)

d-i1 = (d (o-15) - 15 o) / (o-15)

d- i1 = (30 or -30 15 -15 o) / (o-15)

d-i1 = (15 or - 450) / (o- 15)

d-i1 = = (15 or -450) / (o-15)

replace in 1

1 / i2 = 1/20 - (or - 15) / (15 or -450)

1 / i2 = [(15 o-450) - (o-15) 20] / (15 or -150)

1 / i2 = (15 or - 450 - 20 or + 300) / (15 or - 150)

1 / i2 = (-5 or -150) / (15 or -150)

1 / i2 = (or -30) / (3 or - 30)

I2 = 3 (o-10) / (o- 30)

Part B

The height of the image, we use the magnification equation

m = h ’/ h = - i / o

h ’= - h i / o

In our case

h ’= h i2 / o

h ’= h 3 (o-10) / o (o-30)

If they give the distance to the object it is easier

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2 years ago

To practice Tactics Box 9.1 Calculating the Work Done by a Constant Force. Recall that the work W done by a constant force F⃗ at

insens350 [35]

Answer:

The vector magnitudes F and r are always postive, so the sign o W is determined entirely by the angle e between the force and the displacement.Submit Figure 1 off 1 part C

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Energy flows from the sun to _______ to consumers and eventually to _______

Kay [80]

Answer:

the answer is c. producers, detrivores

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A long non-conducting hollow cylinder with inner radius, ri and outer radius, r2 is charged with a uniform positive charge with

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Answer:yup

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