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Genrish500 [490]

2 years ago

5

A graduated cylinder has 50ml of water in it before A marble is dropped in after the marble is dropped into the graduated cylind

er the volume level reads 100ml what is the volume of the marble.

1 answer:

Vikentia [17]2 years ago

5 0

Answer:

0.05L

Explanation:

if i understqnd this correctly the water line should increase with the mass of the marble

100-50=50ml

=0.05L

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How do you transfer kinetic energy to electrical energy using a generator system

san4es73 [151]

Answer: so when a turbine converts the K.E and the potential of any moving fluid (more likely liquid or gas) to energy. once the proc is started the turbine generato, the fluid such as water, steam, combus gasses, or air pushes s big series of blades that have mounted on a shaft, which then will rotate the shaft that’s conn to the generator

Explanation: hope this helped plz mark brainest

3 0

3 years ago

Read 2 more answers

SVEN [57.7K]

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: $f=\frac{v}{\lambda}$

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

$f=\frac{340}{0.22} \\\\f=1545.454...$

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

4 0

3 years ago

A 6.0 kg metal ball moving at 4.0 m/s hits a 6.0 kg ball of putty at rest and sticks to it. The two go on at 2.0 m/s.

lutik1710 [3]

The metal ball lost energy while the putty ball gained energy.

<h3>What is momentum?</h3>

Momentum is the product of mass and velocity of the body. We must note that momentum before collision is equal to momentum after collision.

1) Kinetic energy before collision = 1/2mv^2 = 0.5 * 6 * 4 = 12 J

2) kinetic energy after collision = 0.5 * 6 * 2= 6 J

3) Kinetic energy of putty ball = 0.5 * 6 * 2= 6 J

4) Energy lost by the metal ball = 12 J - 6 J = 6 J

5) Energy gained by the putty ball = 6 J - 0J = 6 J

6) The rest of the energy was converted to heat after the collision.

Learn more about kinetic energy: brainly.com/question/999862

3 0

2 years ago

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately <u>91.93 N</u>

y-component of the tension in the rope on the right is approximately <u>71.135 N</u>

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0

2 years ago

A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0

3 years ago