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3 years ago

What can be defined as the rate at which velocity changes

Physics

2 answers:

siniylev [52]3 years ago

8 0

The rate at which velocity changes can be defined as acceleration or average acceleration.

Explanation:

The acceleration is defined as the rate of change of velocity with respect to time. So, the change in velocity from the initial velocity at different time intervals will help to attain the acceleration attained by the object.

$\ a =\frac{v_{f}-v_{i}}{\Delta t}$

where,

$v_{f} \text { and } v_{i}$ – final and initial velocities, $\Delta t$ – time

If the change in velocity is constant for same time interval, then the acceleration will be constant for that object. So, an object can attain uniform acceleration when the velocity changes uniformly with time.

taurus [48]3 years ago

4 0

Answer:

acceleration

Explanation:

The rate at which velocity changes is the definition of the physical quantity called acceleration, and it is given by the formula: $a=\frac{v_f-v_i}{\Delta t}$

where $\Delta t$ is the time that took to change from the initial velocity $v_i$ to the final velocity $v_f$

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Convert the speed of light, 3*10^8 meters/second (m/s), to miles/day (mi/day).

Sati [7]

I believe that would be 1,080,000,000 Kilometers per hour- lmk if that helps

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3 years ago

Read 2 more answers

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

8 0

4 years ago

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J = 120 J decrease

The missing energy has turned into interned energy in the completely inelastic collision

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4 years ago

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together t

Lina20 [59]

Answer:

$dU=-4.36*10^{-18}J$

Explanation:

From the question we are told that:

Average distance $d_{avg} =5.29*10^{-11}m$

Generally the equation for change in electric potential energy is mathematically given by

$dU=u_f-U_1$

Where

U_1=0 Because of initial lengthy distance apart

And

$U_f=\frac{kq_eq_p}{d}$

$U_f=\frac{9*10^9*1.6*10^{-19}*-1.6*10^{-19}}{5.29*10^{-11}}$

$U_f=-4.36*10^{-18}J$

Therefore

$dU=u_f-U_1$

$dU=-4.36*10^{-18}J-0$

$dU=-4.36*10^{-18}J$

3 0

3 years ago

To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad

Julli [10]

The magnitude of the forces acting at the top are;

$\mathbf{F_{Top, \ x}}$ = 132.95 N

$\mathbf{F_{Top, \ y}}$ = 0

The magnitude of the forces acting at the bottom are;

$\mathbf{F_{Bottom, \ x}}$ = $\mathbf{ F_f}$ = -132.95 N

$\mathbf{F_{Bottom, \ y}}$ = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, B gives;

$\sum M_B$ = 0

Therefore;

$\sum M_{BCW}$ = $\sum M_{BCCW}$

Where;

$\sum M_{BCW}$ = The sum of clockwise moments about B

$\sum M_{BCCW}$ = The sum of counterclockwise moments about B

Therefore, we have;

$\sum M_{BCW}$ = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

$\sum M_{BCCW}$ = $F_R$ × √(6² - 2²)

Therefore, we get;

2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = $F_R$ × √(6² - 2²)

$F_R$ = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, $F_R$ ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, $F_R$ = The magnitude of the frictional force of bottom of the ladder on the floor, $F_f$ but opposite in direction