True or false according to newtons 1st law an unbalanced force can change the motion of an object

Find the change in kinetic energy of a 1.0 kg fish leaping to the right at 12.0 m/s.

sp2606 [1]

Answer:

6J

Explanation:

Given parameters:

Mass of fish = 1kg

Velocity = 12m/s

Unknown:

Change in kinetic energy = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically given as:

K.E = $\frac{1}{2}$ m v²

Now, insert the parameters and solve;

K.E = $\frac{1}{2}$ x 1 x 12 = 6J

The change in kinetic energy is 6J

4 0

2 years ago

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how much force is needed to move a brick with a mass of 345-kg a distance of 28m while doing 1008 j of work?

Travka [436]

    Work = (force) x (distance)

   1,008 J = (force) x (28 m)

Divide each side by 28m : (1,008 kg-m²/sec²) / (28 m) = force

   Force = 36 kg-m/s² = 36 Newtons .
      (about 8.1 pounds)

It doesn't matter what that force accomplishes.
It could be moving a brick, lifting a fish, or pushing a little red wagon.
In order to do 1,008 joules of work in 28 meters, it takes 36 N of force,
in the direction of the 28 meters.

5 0

2 years ago

Sebutkan aplikasi gerak osilasi fisis

VLD [36.1K]

........................................................................................

6 0

2 years ago

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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re

lapo4ka [179]

Answer:

$K_{system} = \frac{k}{10}$

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})$

Here, n = 10

Spring constant of each spring = k

Thus,

$\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})$

$\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})$

$\frac{1}{K_{system}} = \frac{10}{k}$

$K_{system} = \frac{k}{10}$

7 0

2 years ago

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A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$

The CCD sensor is 35.34 mm from the lens

b) Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}$

$m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm$

The person appears 17.67 mm tall on the sensor

7 0

2 years ago