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2 years ago

What is the acceleration if velocity increases from 10 m/s to 15m/s after travelling a distance of 5 metre

Physics

1 answer:

Semenov [28]2 years ago

8 0

Answer:

a=1.25m/s²

Explanation:

GIVEN DATA

vi=10m/s

vf=15m/s

S=5m

TO FIND

a=?

SOLUTION

by using third equation of motion

2as=(vf)²-(vi)²

2a(5m)=(15m/s)²-(10m/s)²

10m×a=225m²/s²-100m²/s²

10m×a=125m²/s²

a= $\frac{125}{10}$

a=12.5m/s²

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

7 0

2 years ago

What is harmonic motion

Tanzania [10]

Answer: NNOOOOOOOOOOOOOOOOOOONONONO

Explanation: simple harmonic motion, in physics, repetitive movement back and forth through an equilibrium, or central, position, so that the maximum displacement on one side of this position is equal to the maximum displacement on the other side. The time interval of each complete vibration is the same. The force responsible for the motion is always directed toward the equilibrium position and is directly proportional to the distance from it. That is, F = −kx, where F is the force, x is the displacement, and k is a constant. This relation is called Hooke’s law.

A specific example of a simple harmonic oscillator is the vibration of a mass attached to a vertical spring, the other end of which is fixed in a ceiling. At the maximum displacement −x, the spring is under its greatest tension, which forces the mass upward. At the maximum displacement +x, the spring reaches its greatest compression, which forces the mass back downward again. At either position of maximum displacement, the force is greatest and is directed toward the equilibrium position, the velocity (v) of the mass is zero, its acceleration is at a maximum, and the mass changes direction. At the equilibrium position, the velocity is at its maximum and the acceleration (a) has fallen to zero. Simple harmonic motion is characterized by this changing acceleration that always is directed toward the equilibrium position and is proportional to the displacement from the equilibrium position. Furthermore, the interval of time for each complete vibration is constant and does not depend on the size of the maximum displacement. In some form, therefore, simple harmonic motion is at the heart of timekeeping.

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2 years ago

Question 5 of 10

muminat

The correct answer to this question is D

5 0

2 years ago

a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes

kiruha [24]

The angle through which the grinding wheel rotates in the first second = 5300 rad

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second = 5300 rad

Learn more about angular velocity here:-brainly.com/question/6860269

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10 months ago

These questions !plz !! i need help!!!

Nataly_w [17]

(6) Wagon B is at rest so it has no momentum at the start. If v is the velocity of the wagons locked together, then

(140 kg) (15 m/s) = (140 kg + 200 kg) v

==> v ≈ 6.2 m/s

(7) False. If you double the time it takes to perform the same amount of work, then you halve the power output:

E / (2t ) = 1/2 × E/t = 1/2 P

3 0

3 years ago

What is the acceleration if velocity increases from 10 m/s to 15m/s after travelling a distance of 5 metre​

What is the acceleration if velocity increases from 10 m/s to 15m/s after travelling a distance of 5 metre