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Anvisha [2.4K]
3 years ago
6

Un movil aumenta su velocidad de 10m/s a 20m/s acelerando uniformemente a razon de 5m/s2 ¿que distancia logro en dicha operacion

?
Physics
1 answer:
Kruka [31]3 years ago
8 0

v₀ = initial velocity of the mobile = 10 m/s

v = final velocity of the mobile = 20 m/s

a = acceleration of the mobile = 5 m/s²

d = distance traveled during this operation = ?

Using the kinematics equation

v² = v²₀ + 2 a d

inserting the above values in the equation

20² = 10² + 2 (5) d

400 = 100 + 10 d

subtracting 100 both side

400 - 100 = 100 - 100 + 10 d

300 = 10 d

dividing both side by 10

300/10 = 10 d/10

d = 30 m

hence mobile travels 30 m.

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Read 2 more answers
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
Dimas [21]
A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
5 0
3 years ago
when 82.5 calories of heat are given to a metallic rod of mass 150g its temperature raises from 20 degree celcius to 25 degree C
marishachu [46]

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

<h3>What is specific heat?</h3>

It is the heat required to raise the temperature of the unit mass of a given substance by a given amount (usually one degree).

A metallic rod of mass 150 g (m) absorbs 82.5 cal of heat (Q) and its temperature raises from 20 °C to 25 °C. We can calculate the specific heat (c) of the metal using the following expression.

Q = c × m × ΔT

c = Q / m × ΔT

c = 82.5 cal / 150 g × (25 °C - 20 °C) = 0.11 cal/g.°C

A 150-g metallic rod with a specific heat of 0.11 cal/g.°C absorbs 82.5 calories of heat and its temperature increases from 20 °C to 25 °C.

Learn more about specific heat here: brainly.com/question/21406849

#SPJ1

8 0
1 year ago
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3
Pachacha [2.7K]

This question is incomplete, the complete question is;

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is

|FI = |QQ'I / d²

where K = 1/4π∈0, and

∈0 = 8.854 × 10⁻¹² C²/(N.m²) is the permittivity of free space.

Consider two point charges located on the x-axis:

one charge, q₁ = -18.5 nC, is located at

x₁ = -1.715m; the charge q₂ = 30.5 nC, is at the origin ( x₂=0 )

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q₃ = 51.0 nC placed between q₁ and q₂ at x₃ = -1.085 m ?

Answer: (Fnet3)x = -3.3287 × 10⁻⁵ N

Explanation:

Given that;

Q₁ = -18.5 nC       Q₃ = 51 nC        Q₂ = 30.5 nC

x₁ = - 1.715m         x₃ = - 1.085m     x₂ = 0

Now

x - component of Net force on charge Q₃ is

(Fnet3)x = -K|Q₁I|Q₃I / r₁3² - -K|Q₂I|Q₃I / r₂3²

(Fnet3)x = -(9×10⁹)(51×10⁻⁹) [ 18.5 / ((-1.085 + 1.715)²) + (30.5 / (-1.085)² ] × 10⁻⁹

(Fnet3)x = -3.3287 × 10⁻⁵ N

6 0
3 years ago
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