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Flauer [41]
4 years ago
15

Under what condition is the instantaneous acceleration of a moving body equal to its average acceleration over time?

Physics
1 answer:
Rzqust [24]4 years ago
3 0
If the acceleration is constant (negative or positive) the instantaneous acceleration cannot be

Average acceleration: [final velocity - initial velocity ] /Δ time

Instantaneous acceleration = d V / dt =slope of the velocity vs t graph

If acceleration is increasing, the slope of the curve at one moment will be higher than the average acceleration.

If acceleration is decreasing, the slope of the curve at one moment will be lower than the average acceleration.

If acceleration is constant, the acceleration at any moment is the same, then only at constant accelerations, the instantaneuos acceleration is the same than the average acceleration.

Constant zero acceleration is a particular case of constant acceleration, so at constant zero acceleration  the instantaneous accelerations is the same than the average acceleration: zero. But, it is not true that only at zero acceleration the instantaneous acceleration is equal than the average acceleration.

That is why the only true option and the answer  is the option D. only at constant accelerations.
You might be interested in
Whats the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s
Vesnalui [34]

Answer:

600J

Explanation:

1/2mv^2

1/2(12)(10^2)

6(100)

600

7 0
4 years ago
If an equation is dimensionally correct is thar equation a right equation<br>​
Nady [450]

If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true. Dimensional analysis is a technique used to check whether a relationship is correct

8 0
3 years ago
A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about
Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity  Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[0^{2} -11^{2}]

x-x₀=6.05*10^{-2} m

       =6.05 cm

3 0
3 years ago
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 514 N. As the drawing sh
n200080 [17]

Answer:

Part a)

T_L = 155.4 N

Part b)

T_R = 379 N

Explanation:

As we know that mountain climber is at rest so net force on it must be zero

So we will have force balance in X direction

T_L cos65 = T_R cos80

T_L = 0.41 T_R

now we will have force balance in Y direction

mg = T_L sin65 + T_Rsin80

514 = 0.906T_L + 0.985T_R

Part a)

so from above equations we have

514 = 0.906T_L + 0.985(\frac{T_L}{0.41})

514 = 3.3 T_L

T_L = 155.4 N

Part b)

Now for tension in right string we will have

T_R = \frac{T_L}{0.41}

T_R = 379 N

3 0
3 years ago
Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
NISA [10]

Answer:

Fnet = F√2

Fnet = kq²/r² √2

Explanation:

A exerts a force F on B, and C exerts an equal force F on B perpendicular to that.  The net force can be found with Pythagorean theorem:

Fnet = √(F² + F²)

Fnet = F√2

The force between two charges particles is:

F = k q₁ q₂ / r²

where

k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.

If we say the charge of each particle is q, then:

F = kq²/r²

Substituting:

Fnet = kq²/r² √2

5 0
3 years ago
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