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gladu [14]
3 years ago
9

You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after

pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Physics
1 answer:
kotykmax [81]3 years ago
7 0

Answer: d = 33 cm or 0.33 m

Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:

W = F.d.cosθ

F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.

For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:

W = F.d.cos(0)

W = F.d

To determine d:

d = \frac{W}{F}

d = \frac{3}{9}

d = 0.33 m

The distance d the block ice moved is 33 cm.

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In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.

This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units. 
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3 years ago
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A 1-kg iron frying pan is placed on a stove. The pan increases from 20°C to 250°C. If the same amount of heat is added to a pan
sergejj [24]

Here mass of the iron pan is given as 1 kg

now let say its specific heat capacity is given as "s"

also its temperature rise is given from 20 degree C to 250 degree C

so heat required to change its temperature will be given as

Q = ms \Delta T

Q = 1*s*(250 - 20)

Q = 1*s*230

now if we give same amount of heat to another pan of greater specific heat

so let say the specific heat of another pan is s'

now the increase in temperature of another pan will be given as

Q = ms'\Delta T

1*s*230 = 1* s' * \Delta T

now we have

\Delta T = (\frac{s}{s'})*230

now as we know that s' is more than s so the ratio of s and s' will be less than 1

And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature

So correct answer is

<u>A) The second pan would reach a lower temperature.</u>

3 0
3 years ago
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A car is going 8 meters per second on an access road into a highway
TiliK225 [7]

Answer:

20.96 m/s^2 (or 21)

Explanation:

Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.

At first, we know a car is going 8 m/s, that is its initial velocity.

Then, we know the acceleration, which is 1.8 m/s/s

We also know the time, 7.2 second.

Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.

(final velocity - initial velocity) = time * acceleration

final velocity = time*acceleration + initial velocity

After plugging the found values in, we get 20.96 m/s/s, or 21 m/s

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3 years ago
A bird sits on a high-voltage power line with its feet 3.87 cm apart. The wire is made from aluminum, is 2.11 cm in diameter, an
Svetlanka [38]

Answer:

ΔV=0.484mV

Explanation:

The potential difference across the end of conductor that obeys Ohms law:

ΔV=IR

Where I is current

R is resistance

The resistance of a cylindrical conductor is related to its resistivity p,Length L and cross section area A

R=(pL)/A

Given data

Length L=3.87 cm =0.0387m

Diameter d=2.11 cm =0.0211 m

Current I=165 A

Resistivity of aluminum p=2.65×10⁻⁸ ohms

So

ΔV=IR

=I(\frac{pL}{A})\\ =I(\frac{pL}{\pi r^{2} } )\\=I(\frac{pL}{\pi (d/2)^{2} } )\\=165A((\frac{(2.65*10^{-8})(0.0387m)}{\pi (0.0211m/2)^{2} } ))\\=4.84*10^{-4}V

ΔV=0.484mV  

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3 years ago
Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint
djyliett [7]

Answer:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

Explanation:

For this case the figure attached shows the illustration for the problem

We have an inverse square law with distance for the force, so then the force of gravity between Earth and the spaceship is lower when the spaceship is far away from Earth.

Th formula is given by:

F = G \frac{m_{Earth} m_{Spaceship}}{r^2}

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m_{Earth} represent the mass for the earth

m_{spaceship} represent the mass for the spaceship

r represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

F_5 >F_4>F_1 >F_2>F_3

Where F_i represent the force for each of the 5 cases i -1,2,3,4,5 presented on the figure attached.

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