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pickupchik [31]
3 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]3 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]3 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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A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
mr_godi [17]

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

5 0
3 years ago
Your teacher pokes a hole in a paper cup and allows the water to drain. She then
Pavel [41]

Answer:

Explanation:v

2

h = 2g (H-h)

5 0
2 years ago
In which part of the EM spectrum do you see the colors red, orange, yellow, green, blue, indigo, and violet (ROYGBIV);
Lilit [14]

Answer:

gamma rays

Explanation:

hope it helps

7 0
2 years ago
before colliding the momentum of block A is -100 kg*m/s, and block B is -150 kg*m/s. after, block A has a momentum -200 kg*m/s.
Virty [35]

The momentum of block B after the collision is -50 kg m/s.

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:

p_A + p_B = p'_A + p'_B

where:

p_A = -100 kg m/s is the momentum of block A before the collision

p_B = -150 kg m/s is the momentum of block B before the collision

p'_A = -200 kg m/s is the momentum of block A after the collision

p'_B is the momentum of block B after the collision

Solving for p'_B, we find:

p'_B = p_A + p_B - p'_A = -100 +(-150) -  (-200)=-50 kg m/s

So, the momentum of block B after the collision is -50 kg m/s.

Learn more about momentum:

brainly.com/question/7973509

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6 0
3 years ago
A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coeffici
Ivenika [448]

Answer:

Coefficient of friction will be 0.296

Explanation:

We have given initial speed of the stone u = 8 m /sec

It comes to rest so final speed v = 0 m /sec

Distance traveled before coming to rest s = 11 m

According to third equation of motion

v^2=u^2+2as

So 0^2=8^2+2\times a\times 11

a=\frac{-64}{22}=-2.90m/sec^2

Acceleration due to gravity g=9.8m/sec^2

We know that acceleration is given by

a=\mu g

So 2.90=9.8\times \mu \\

\mu =\frac{2.9}{9.8}=0.296

So coefficient of friction will be 0.296

3 0
3 years ago
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