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3 years ago

Can you use a machine to gain both force and speed at the same time? explain.

1 answer:

8 0

Well, It rather depends on your definition of "machine." The normal physics set of simple machines - levers, pulleys, ramps all give you increased the force at the expense of reduced speed or increased the rate at the cost of reduced force. So, no - by definition a machine is an arrangement for multiplying one while paying the cost by reducing the other. You are looking at an example of the Conservation of Energy. One of the giant rules we are pretty sure cannot be violated.<span>
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If a 10kg block is at rest on a table and a 1200N force is applied in the eastward direction for 10 seconds, what is the acceler

gavmur [86]

Answer:

a = 120 m/s²

Explanation:

We apply Newton's second law in the x direction:

∑Fₓ = m*a Formula (1)

Known data

Where:

∑Fₓ: Algebraic sum of forces in the x direction

F: Force in Newtons (N)

m: mass (kg)

a: acceleration of the block (m/s²)

F = 1200N

m = 10 kg

Problem development

We replace the known data in formula (1)

1200 = 10*a

a = 1200/10

a = 120 m/s²

6 0

3 years ago

A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to

ruslelena [56]

T1=40°C=313K
T_2=-10°C=263K

Applying Charles law

$\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}$

$\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}$

$\\ \sf\Rrightarrow V_2=84.02ml$

6 0

1 year ago

A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?

Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration = 24m/s²

Unknown:

Force needed by the cheetah to run = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

Force = mass x acceleration

Insert the given parameters and solve;

Force = 12 x 24 = 288N

7 0

2 years ago

two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have: