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Gekata [30.6K]
3 years ago
14

A car moved 40 km north and 90 km south. What is the displacement of the car?

Physics
1 answer:
il63 [147K]3 years ago
8 0

Answer:

A. 50 km south

Explanation:

hope this helps!

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the force of friction between a 1000 kg car and the road is 10000 N, what is the fastest acceleration the car can achieve?​
nordsb [41]

Answer:

1000

Explanation:

7 0
3 years ago
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A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and rea
Ugo [173]

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement  a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t  Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s  , t= 30s

θ = (58 rad/s)*(30)

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt  = 290 rad+ 1740 rad + 1201.42 rad

θt  = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

7 0
3 years ago
Velocity can best be described by which of the following statements?
pogonyaev

Answer: Velocity can best be described as, the speed in a given direction.

Explanation: To find the answer, we need to know more about the Velocity of a body.

<h3>What is Velocity of a body?</h3>
  • Velocity is the rate of change of displacement.
  • It's a vector quantity and is measured in m/s.
  • It can be positive, negative or zero.
  • A body is said to be in uniform motion, then its velocity remains constant.
  • Change in velocity can be a change in speed.
  • The magnitude of velocity is less than or equal to speed.

Thus, we can conclude that, the option C is best describing velocity.

Learn more about velocity here:

brainly.com/question/28108466

#SPJ4

8 0
2 years ago
Read 2 more answers
A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
C, N, Ne, Ar which contains a metal, nonmetal, noble gas , metalloid
postnew [5]

Answer:

c,carbono nonmetal

n, nitrogen nonmetal

ne, neón noble gas

ar,argon noble gas

4 0
3 years ago
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