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Alchen [17]
3 years ago
6

Which is not expecting acceleration?

Physics
2 answers:
TiliK225 [7]3 years ago
7 0

#1)

1).  A point on the outside of a spinning top whose rotational speed is constant.

It will not accelerate as its net speed is ZERO

2).  A skydiver whose air resistance is equal to that of her weight.

Since weight is counterbalanced by resistance so here net force is zero so acceleration must be zero here

3).   A car on the freeway experiencing a net force of -120 N.

Since there it acts net force F = -120 N so it will accelerate here

4).  A submerged beach ball whose buoyant force is eight times the force of gravity on it.

Since buoyant force is more than weight of the ball so it will accelerate upwards

#2)

A non accelerating fra me is always a valid inertial frame

A frictionless spinning merry go round.

(since it is having centripetal acceleration so it is non inertial frame)

A falling rock.

(Since it is falling with acceleration a = 9.8 m/s^2 so it is non inertial)

A hot air balloon moving at 30 degrees east of north with no net force.

(Since net force on hot air balloon is ZERO so it is Inertial Frame)

A space shuttle whose boosters just ignited for takeoff.

(Since space shuttle ignite the boosters so it will accelerate here and hence it is non inertial)

#3

here two forces given as

F_1 = 4320 N towards right

F_2 = 4380 N towards Left

Since left side applied force is more than right side applied force so net force on it will be towards Left and it is given by

F = 4380 - 4320 = 60 N

So it is 60 N Left

Mnenie [13.5K]3 years ago
4 0

Answer:

1. A submerged beach ball whose buoyant force is eight times the force of gravity on it.

2. A hot air balloon moving at 30 degrees east of north with no net force.

3. 60 N to the left

Explanation:

1. A submerged ball is not expecting any form of acceleration. This is because there is no form of motion on it. The only forces that are experienced on the ball are the compressing forces from the water.

2. A hot air balloon will be a good point of reference. This is because it has no definite direction of movement or acceleration.

3. The sum of the forces will be:

3280 - 3220 = 60

so the resultant force will be 60 N to the left.

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You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
4 years ago
List 5 possible effects of not adhering to standards of measurement
murzikaleks [220]

Answer:

factual evidence of customer-service levels.

better understanding of cross-functional performance.

enhanced alignment of operations with strategy.

evidence-based determination of process improvement priorities.

detection of performance trends.

better understanding of the capability range of a process.

3 0
3 years ago
In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.
Andru [333]

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

<u>Given the following data;</u>

  • Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

<u>Conversion:</u>

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

<em>36.35 g = 36.35E15 micrometer</em>

II. Millimetre

<u>Conversion:</u>

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

<em>36.35 g = 363.5 millimetre</em>

III. Kilogram

<u>Conversion:</u>

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

<em>36.35 g = 0.03635 kilogram</em>

<u>Note:</u>

  • g is the symbol for grams.
  • Exp (E) means exponential = 10
  • um is the symbol for micrometer.
4 0
3 years ago
2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg &amp; object B’s mass is 11kg. After the colli
ss7ja [257]

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

3 0
3 years ago
How many oxygen atoms are used to bond with sulfur in the compound "sulfur dioxide"?
kherson [118]
2 oxygen atoms are used to bond with sulfer
7 0
3 years ago
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