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scZoUnD [109]
3 years ago
11

Does uniform circular motion have constant acceleration?

Physics
1 answer:
AysviL [449]3 years ago
7 0
Yes. It's (speed squared)/(radius of the circle).
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The Sun appears to move across the sky during the day. The best explanation for this apparent motion is that Earth is
Harrizon [31]

Answer:

C

Explanation:

5 0
3 years ago
Read 2 more answers
A 2137 kg car moving east at 12.91 m/s collides with a 3264 kg car moving north. The cars stick together and move as a unit afte
IRINA_888 [86]

Answer:

The speed of the 3264 kg car before collision is 6.32 m/s

Explanation:

Let first car, mass m₁, moving east be moving with velocity, v₁ and second car, mass m₂, moving north be moving with velocity, v₂. Let the velocity of the two cars after collision be v₃

In a Collison (whether elastic or inelastic), momentum is always conserved.

Momentum before collision = momentum after collision.

Momentum before collision = m₁v₁ + m₂v₂

v₁ = (12.91î) m/s in vector form, m₁ = 2137 kg

v₂ = ?, v₂ = (v₂j) m/s in vector form, m₂ = 3264 kg

Momentum before collision = (2137)(12.91î) + (3264)(v₂j) = (27590î + 3264v₂j) kgm/s

Momentum after collision = (total mass of the cars after collision) × (velocity of the stuck-together cars after collision)

Total mass of the cars after collision = m₁ + m₂ = 2137 + 3264 = 5401 kg

Velocity after collision, v₃ = 6.38 m/s In the N36.8°E direction, put in vector form,

v₃ = (6.38 cos 36.8°î + 6.38 sin 36.8°j)

v₃ = (5.11î + 3.82j) m/s

Momentum after collision = 5401 (5.11î + 3.82j) = (27590î + 20641.4j) kgm/s

Momentum before collision = momentum after collision.

(27590î + 3264v₂j) = (27590î + 20641.4j)

Comparing components

3264v₂ = 20641.4j

v₂ = 6.32 m/s

Hope this Helps!!!

6 0
3 years ago
A 1500-kg car locks its brakes and skids to a stop on a slippery horizontal road, leaving skid marks that are 15 m long. How muc
Harman [31]

Answer:

E=88200\ J

Explanation:

Given:

  • mass of car, m=1500\ kg
  • distance of skidding after the application of brakes, d=15\ m
  • coefficient of kinetic friction, \mu_k=0.4

<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

f=\mu_k.N

where N = normal reaction by ground on the car

f=0.4\ties 1500\times 9.8

f=5880\ N

<em>Now from the work-energy equivalence:</em>

E=f.d

E=5880\times 15

E=88200\ J is the dissipated energy.

3 0
3 years ago
A constant force is applied to an object, causing the object to accelerate at 10 m/s^2. What will the acceleration be if: a) The
Liula [17]
What you need to know is that the force is

F=ma

The force is the product of mass and acceleration

this means that the acceleration is

a=F/m

a) The force is halved?
this means that f will be \frac{F}{2} now:

a=\frac{F}{2m}

So the accelaration will also he halved (it's the original acceleratation divided by 2)


 b) The object's mass is halved?
a=\frac{F}{m/2}=a=\frac{F2}{m}

which is the original acceleration times two!! so it will double


c) The force and the object's mass are both halved?
now we have

a=\frac{F/2}{m/2}=a=\frac{2F}{2m}=a=\frac{F}{m}

so they will cancel each other out and the acceleration will stay the same!











5 0
2 years ago
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

t(final) =  \frac{w1 \times t1 + w2 \times t2}{w1 + w2}

T(final) = 500* 10 + 100*70/600 = 20°C

4 0
3 years ago
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