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3 years ago

If you saw a waxing gibbous moon what phase would you expect one week later?

1 answer:

6 0

Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.

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Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

An electric generator transforms mechanical energy into electrical energy. This process could be done by which of these?

Dimas [21]

Answer:

its d

Explanation:

6 0

3 years ago

Read 2 more answers

if an average cloud has a density of 0.5 g/m3 and has a volume of 1,000,000,000 m3,what is the weight of an average cloud

sveta [45]

If you have (1 x 10⁹) cubic meters of volume, and 1/2 gram of mass
in each cubic meter, then you have

(1/2 x 10⁹) grams = (5 x 10⁸) grams = (5 x 10⁵) kilograms of mass .

On Earth, that mass weighs 4,900,000 newtons .

(about 1,102,300 pounds)

(about 551 tons)

3 0

3 years ago

A man pushes on a wall what does Newton third law say must happen

Verdich [7]

If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.

8 0

3 years ago

Read 2 more answers

A wheel 1.0 m in radius rotates with an angular acceleration of 4.0rad/s2 . (a) If the wheel’s initial angular velocity is 2.0 r

Oliga [24]

Answer:

(a) ωf= 42 rad/s

(b) θ = 220 rad

Explanation:

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

There is tangential acceleration (at ) and 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)

at = α*R Formula (3)

v= ω*R Formula (4)

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 velocity ( rad/s)

ωf: final angular velocity ( rad/s)

R : radius of the circular path (cm)

at : tangential acceleration (m/s²)

v : tangential speed (m/s)

Data

α = 4.0 rad/s² : wheel’s angular acceleration

t = 10 s

ω₀ = 2.0 rad/s : wheel’s initial angular velocity

R = 1.0 m : wheel’s radium

(a) Wheel’s angular velocity after 10 s

We replace data in the formula (1):

ωf= ω₀ + α*t

ωf= 2 + (4)*(10)

ωf= 42 rad/s

(b) Angle that rotates the wheel in the 10 s interval

We replace data in the formula (2):

θ= ω₀*t + (1/2)*α*t²

θ= (2)*(10) + (1/2)*(4)*(10)²

θ= 220 rad

θ= 220 rad

(c) Tangential speed and acceleration of a point on the rim of the wheel at the end of the 10-s interval

We replace data in the Formula (3)

at = α*R = (4)(1)

at = 4 m/s²

We replace data in the Formula (4)

v= ω*R = (42)*(1)

v = 42 m/s

6 0

3 years ago