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NARA [144]
3 years ago
5

How can energy crisis be aberted?​

Physics
1 answer:
omeli [17]3 years ago
6 0

Answer:

Try and depend more on renewable energy sources. Use products that are more energy efficient. Make use of lighting control measures. Maintain climate change

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A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.
kykrilka [37]

1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

1)

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

W=K_f - K_i

where

W is the work done

K_f is the final kinetic energy

K_i is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

K_i = 0

Therefore, the final kinetic energy is:

K_f = K_i + W=0+50=50 J

2)

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the cart in this problem, we have:

K = 50 J is its final kinetic energy

m = 10 kg is the cart

Therefore, solving the formula for v, we find its speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s

3)

We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

PE=mgh

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

h is the heigth of the cart

When the cart reaches the maximum height, all the kinetic energy has been converted into potential energy, so we can write:

K=PE\\\frac{1}{2}mv^2=mgh

Re-arranging,

h=\frac{v^2}{2g}

And since we know the initial speed of the cart along the ramp,

v = 3.2 m/s

we can find the maximum height reached along the ramp:

h=\frac{3.2^2}{2(9.8)}=0.5 m

Learn more about work, kinetic energy and potential energy:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/6536722

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0
3 years ago
A violin string has a standard length of 32.8 cm. It sounds the musical note A (440 Hz) when played without fingering. How far f
Rus_ich [418]

Answer:

The finger has to be placed at 5.25cm

Explanation:

The steps and appropriate formula in musical instrument is as shown in the attachment.

5 0
3 years ago
A crane lifts a 1,750 kg mass using a steel cable whose mass per unit length is 0.88 kg/m. What is the speed of transverse waves
Sauron [17]

Answer:

139.6m/s

Explanation:

Calculate the tension first, T=m*g

mass(m): 1750kg, gravity(g): 9.8m/s^2

T= 1750*9.8

 =17150N

Then calculate the wave speed using the equation v = √ (T/μ)

v= √(17150N)/(0.88kg/m)

 =139.6m/s

4 0
3 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kg when weighed in air. The density of
sasho [114]

Answer:

m = 4.5021 kg

Explanation:

given,

Apparent mass of aluminium = 4.5 kg

density of air = 1.29 kg/m³

density of aluminium = 2.7 x 10⁷ kg/m³

true mass of the aluminium = ?

Weight in Vacuum

W = m g

W = ρV g

Air buoyancy acting on aluminium

B = ρ₀V g

Volume is the same in both cases since the volume of the aluminum

displaces an equal amount of volume air.

Apparent weight:

ρV g − ρ₀V g = 4.5 g

ρV − ρ₀V = 4.5

V = \dfrac{4.5}{\rho - \rho_0}

m = ρV

m = \dfrac{4.5\times \rho}{\rho - \rho_0}

m = \dfrac{4.5\times 2700}{2700 - 1.29}

m = 4.5021 kg

4 0
3 years ago
A 26.0-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 11.0 m th
frozen [14]

Answer:

18 m/s

Explanation:

Force is given by mass × acceleration

<em>F = ma</em>

<em>a = F/m</em>

For the first 11.0 m, there is no friction. Hence, the constant force is applied fully on the crate. The acceleration is

<em>a = </em>225/26.0<em> </em>m/s²

By the equation of motion, <em>v² = u² + 2as</em>

where <em>v</em> is the final velocity, <em>u</em> is the initial velocity, <em>a</em> is the acceleration and <em>s</em> is the distance travelled.

Using the values for the first part of the motion,

<em>v² = </em>0² + 2 × 8.65 × 11.0

<em>v = </em>13.8 m/s

This is the initial velocity for the second part of the motion. This part has friction of coefficient 0.20.

The frictional force = 0.20 × weight = 0.20 × 26.0 × 9.8 = 50.96 N

The effective force moving the crate in the second motion = 225 - 50.96 N = 174.04 N

This gives an acceleration of 174.04/26.0 = 6.69 m/s².

Using parameters for the equation of motion, <em>v² = u² + 2as</em>

<em>v² = </em>13.8² + 2 × 6.69 × 10 = 324.24

<em>v = </em>18<em> </em>m/s

8 0
4 years ago
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