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Kaylis [27]
3 years ago
10

Kinetic energy of an object is equal to

Physics
2 answers:
Arte-miy333 [17]3 years ago
7 0
Choice-'b' says the formula for kinetic energy in words.

     KE = (1/2) · (M) · (S²)
DanielleElmas [232]3 years ago
7 0

Answer:

b

Explanation:

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Svetradugi [14.3K]
Erosion, formation of sinkholes, loss of biodiversity, and contamination of soil, groundwater and surface water by chemicals from mining processes.
7 0
3 years ago
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A 4.25 kg block is sent up a ramp inclined at an angle theta=37.5° from the horizontal. It is given an initial velocity ????0=15
wel

Answer:

d = 11.79 m

Explanation:

Known data

m=4.25 kg  : mass of the block

θ =37.5°  :angle θ of the ramp with respect to the horizontal direction

μk= 0.460  : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the block

W: Weight of the block : In vertical direction

N : Normal force : perpendicular to the ramp

f : Friction force: parallel to the ramp

Calculated of the W

W= m*g

W=  4.25 kg* 9.8 m/s² = 41.65 N

x-y weight components

Wx= Wsin θ= 41.65*sin 37.5° = 25.35 N

Wy= Wcos θ =41.65*cos 37.5° =33.04 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 33.04 N

Calculated of the f

f = μk* N= 0.460*33.04

f = 15.2 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

-Wx-f = m*a

 -25.35-15.2 = (4.25)*a

-40.55 =  (4.25)*a

a = (-40.55)/ (4.25)

a = -9.54 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  (m)

v₀: initial speed  (m/s)

vf: final speed   (m/s)

Data:

v₀ = 15 m/s

vf = 0

a = -9.54 m/s²

We replace data in the formula (2)  to calculate the distance along the ramp the block reaches before stopping (d)

vf²=v₀²+2*a*d

0 = (15)²+2*(-9.54)*d

2*(9.54)*d =   (15)²

(19.08)*d = 225

d = 225 / (19.08)

d = 11.79 m

3 0
3 years ago
Which of the following numbers is correctly represented by 4 x 10-7? 0.0000004 40,000,000 0.000004
creativ13 [48]
0.0000004. since it is 10^-7 (note the negative), you would move thedecimal point 7 places to the left.
5 0
3 years ago
When is the magnitude of the acceleration of a mass on a spring at its maximum value?
Andreas93 [3]

Answer:

A. when the mass has a speed of zero

Explanation:

In mass-spring system, the velocity and the acceleration are in anti-phase, which means that when one of the two quantities is maximum, the other one is zero, and vice-versa.

In fact:

- When the displacement of the spring is zero (x=0), the velocity is maximum, due to conservation of energy. In fact, as the displacement is zero, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is zero, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be maximum, and so the velocity (v) is also maximum. On the cotnrary, acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=0, then the force is zero: F=0, and so the acceleration is zero as well.

- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is maximum, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=maximum, then the force is maximum, and so the acceleration is maximum as well.

Based on this, the correct answer is

A. when the mass has a speed of zero


5 0
3 years ago
Read 2 more answers
An emf of 28.0 mV is induced in a 501 turn coil when the current is changing at a rate of 12.0 A/s. What is the magnetic flux th
dmitriy555 [2]

Answer:

Φ = 5.589×10⁻⁵  Wb

Explanation:

The inductance of a coil is given as

L = e/(di/dt) ..................... Equation 1

Where L = inductance of the coil, e = induced e.m.f, di/dt = rate of change of current in the coil.

Also,

The inductance of each turn of the coil when a magnetic field is step up in the coil  is

L = NΦ/i ................. Equation 2

Where N = number of turns, Φ = magnetic field, i = current.

equating equation 1 and equation 2

e/(di/dt) = NΦ/i

making Φ the subject of the equation,

Φ = (e×i)/N.(di/dt) .................. Equation 3

Given: e = 28.0 mV = 0.028 V, N = 501 turns, di/dt = 12.0 A/s, i = 4.00 a

Substitute into equation 3,

Φ = (0.028×4)/(12×501)

Φ = 0.112/2004

Φ = 5.589×10⁻⁵ Weber

Φ = 5.589×10⁻⁵ Wb

6 0
3 years ago
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