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pickupchik [31]
2 years ago
9

tas watches as his uncle changes a flat tire on a car. his uncle raises the car using a machine called a jack. each time his unc

le pushes down on the jack handle, the car rises up. tas sketches the jack and labels it using the symbols f, to represent force, and d, to represent distance. he uses large and small letters to compare the sizes of the forces and distances. which set of labels is correct for the side of the jack that tas’s uncle pushes on? a large f and a small d a small f and a small d a large f and a large d a small f and a large d
Physics
2 answers:
Alenkinab [10]2 years ago
8 0
Small f and large L.

People needs help of machines to increase their force.

People cannot lift a car without a machine.

Using the leverage  or hydraulic principles the machines increase your force.

If you use a large leverage you execute a large movement with little force and as result the ohter side will move small distances with a greater force.

I hope this help. Please, let me know.
kakasveta [241]2 years ago
7 0

The answer is

-Small f and large D.

The explanation:

-when The car jack is an example of a machine, which is defined as anything that a person can use to make the exertion of force easier.

-So with the small force he exerts on the jack, the distance that the car is lifted up increases .

and People needs help of machines to increase their force.  People cannot lift a car without a machine. Using the leverage  or hydraulic principles the machines increase your force.

If you use a large leverage you execute a large movement with little force and as result the other side will move small distances with a greater force.

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3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
andreev551 [17]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

so we get as,

W(output) = (22/100) x 2.2 x 10^7

=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7

hence, W(output) = 4.84 x 10^6 J

The useful work done on the mass is 4.84 x 10^6 J

5 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m
r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0
2 years ago
A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t
Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s

Time taken to reach maximum height is 0.97 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

v=u+at\\\Rightarrow v=0+9.81\times 86.47 = 848.27\ m/s

Speed just before hitting the water is 848.27 m/s

3 0
2 years ago
When 0.1523 g of liquid pentane (CH) combusts in a bomb calorimeter, the temperature rises from 23.7C to 29.8 C. What is U for t
pashok25 [27]

Answer:

U for the reaction is 15,048.58 kJ/mol of pentane.

Explanation:

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Moles of pentane = mass/MW

Mass = 0.1523 g

MW of pentane (C5H12) = 72 g/mol

Moles of pentane = 0.1523/72 = 0.00212 mol

U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol

8 0
2 years ago
In which of the following examples does the object have both kinetic and potential energy? Select all that apply.
notsponge [240]
I believe the answer is H for when you bounce it, it has stress when it hits the floor and then goes up giving it kinetic
6 0
2 years ago
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