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Whitepunk [10]
3 years ago
8

Different between pulling and pushing force​

Physics
1 answer:
yuradex [85]3 years ago
4 0

Answer:

Push and pull are both forces, but the difference is in their direction at which it is applied. If the force applied in the direction of motion of the particle then we call it push. If that force applied in the direction OPPOSITE to the motion of the particle then it it termed as pull

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Which statement best describes a similarity between presidential and
zlopas [31]

The statement 'both allow citizens to vote for members of the legislature' describes a similarity between presidential and parliamentary democracies. They are different forms of government.

<h3>Presidential and parliamentary democracies</h3>

Democracy refers to a form of government where people vote to choose authorities designed to decide legislation.

A presidential democracy is when the presented is voted to be the head of government (president) separately from the legislative system.

Parliamentary democracy is a form of government where the political party having the greatest representation in the legislature forms the government.

Learn more about presidential and parliamentary democracies here:

brainly.com/question/3069300

8 0
3 years ago
What is the change in internal energy if 20 J of heat is released from a system and the system does 50 J of work on the surround
hjlf

The correct answer is a -80 J

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3 years ago
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Question number 11 how did we found the answer ?
BaLLatris [955]

Answer:

Option A. 57.14 Ω

Explanation:

From the question given above, the following data were obtained:

Resistor 1 (R₁) = 100 Ω

Resistor 2 (R₂) = 400 Ω

Resistor 3 (R₃) = 200 Ω

Equivalent Resistor (Rₚ) =?

The equivalent resistor in the above circuit can be obtained as follow:

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

1/Rₚ = 1/100 + 1/400 + 1/200

Find the least common multiple (lcm) of 100, 400 and 200. The result is 400. Divide 400 by 100, 200 and 400 respectively and multiply the result with the numerator as shown

1/Rₚ = (4 + 1 + 2)/400

1/Rₚ = 7/400

Invert

Rₚ = 400/7

Rₚ = 57.14 Ω

6 0
3 years ago
PHYSICS PLEASE HELP 80 PTS!!!!!!
VladimirAG [237]

1.  B) 32 m/s

The speed of the baseball is given by

v=\frac{d}{t}

where

d = 48 m is the distance travelled

t = 1.5 s is the time taken

Substituting,

v=\frac{48}{1.5}=32 m/s

2. C) 12 m/s

The average velocity is given by

v=\frac{d}{t}

where

d = 60 - 0 = 60 m is the displacement

t = 5 s is the time interval

Substituting,

v=\frac{60}{5}=12 m/s

3. Missing diagram

4. D) velocity

In fact, velocity is defined as the ratio between the displacement (\Delta s) and the time interval (\Delta t) needed to achieve that change in position:

v=\frac{\Delta s}{\Delta t}

Similarly, acceleration is defined as the ratio between the change in velocity (\Delta v) and the time interval (\Delta t):

a=\frac{\Delta v}{\Delta t}

Comparing the two equations, we see that displacement is to velocity as velocity is to acceleration.

5. C) 5/1 cm/year

We know that the ridge moves 25 cm in 5 years, so we have:

- Displacement (d): d = 25 cm

- Time interval (t): t = 5 y

Using the equation for the velocity:

v=\frac{d}{t}=\frac{25 cm}{5 y}=5 cm/y

6.  D

The missing graph is attached. Each graph represents a distance (on the y-axis, measured in metres) versus a time (on the x-axis, measured in seconds). Therefore, the speed in each graph can be simply calculated from the slope of the curve, since speed is the ratio between distance and time:

v=\frac{\Delta y}{\Delta x}

For each graph:

A. \frac{\Delta y}{\Delta x}=\frac{8.6-2}{10}\sim 0.66

B. \frac{\Delta y}{\Delta x}=\frac{5.7-0}{10}\sim 0.57

C. \frac{\Delta y}{\Delta x}=\frac{9.6-3}{10}\sim 0.66

D. \frac{\Delta y}{\Delta x}=\frac{7.5-0}{5}\sim 1.5

So we can say that the correct graph must be graph D, the closest to 1.57.

7. B) The car has come to a stop and has zero velocity

The missing graph is attached.

The graph shows the position of the car versus time. From the graph, we can see that in segment B, the position of the car does not change: this means that its velocity is zero, since the velocity is defined as the ratio between the change in position (displacement) and the time interval:

v=\frac{\Delta s}{\Delta t}

And since the displacement is zero, the velocity is also zero.

8.  C) from 4.5 to 5.0 seconds

Missing graph is attached.

The graph represents the distance covered versus the time taken: therefore, the speed of the ball can be evaluated from the slope of the curve.

The only region in which the slope of the curve is zero is between 4.5 seconds and 5.0 seconds: therefore, this is the region where the soccer ball is not moving.

9. D) diagonal line with varying slope, from 3 to 4.

The table of data is:

Time (s)  0 1 2 3 4

Distance (m) 0 3 6 12 16

We want to know how the distance/time graph would like like. We have:

- At t = 1, the distance is d=3, so the slope is \frac{d}{t}=\frac{3}{1}=3

- Similarly, at t = 2:  \frac{d}{t}=\frac{6}{2}=3

- Instead, at t =3, the slope is \frac{d}{t}=\frac{12}{3}=4

- Similarly, at t=4,\frac{d}{t}=\frac{16}{4}=4

So the correct answer is D.

10. D) To move in a circle requires an acceleration and therefore a net force. This force is supplied by the slingshot.

For an object in circular motion, the velocity constantly changes (because the direction changes): this means that an acceleration is required (towards the centre of the circle), and therefore a force must be exerted (also towards the centre of the circle) to keep the object in the circular path.

As soon as this force is removed, the object is "free" to leave the circular trajectory and continue its motion following a straight path at constant velocity, according to Newton's first law.

11. B) The acceleration will become 1/4 as much.

Re-arranging Newton's second law formula,

F=ma \rightarrow a = \frac{F}{m} (1)

The force exerted on the satellite is actually the force of gravity:

F=\frac{Gm M}{d^2} (2)

Substituting (2) into (1),

a=\frac{F}{m}=\frac{GM}{d^2}

We see that there is no dependance on the mass of the satellite (m), but only on the distance. Since in this problem the distance is doubled (d'=2d), the new acceleration will be:

a'=\frac{GM}{(2d)^2}=\frac{1}{4}(\frac{GM}{d^2})=\frac{a}{4}

12. C) divide: distance : velocity

Velocity is distance divided by time:

v=\frac{d}{t}

Multiplying by t on both sides,

v\cdot t = \frac{d}{t} \cdot t \rightarrow vt = d

and dividing by v on both sides,

\frac{vt}{v}=\frac{d}{v} \rightarrow t = \frac{d}{v}

13. A) -8 km/h

Let's call:

v_k = +5 km/h the velocity of the kite

v_p = -3 km/h your velocity (opposite direction)

In your frame of reference, the velocity of the kite will be:

v'_k = v_p - v_k = -3 -(+5) = -8 km/h

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4 years ago
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Como se realiza la salida baja en el atletismo?
goldfiish [28.3K]

Answer:

Honestamente yo no se cual es la pregunta que escribiste

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