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Svetllana [295]
2 years ago
9

What's faster a slow car or a fast feather?​

Physics
2 answers:
krok68 [10]2 years ago
5 0

Answer:

A fast feather

Explanation:

The faster any item is, the more momentum it has

zhuklara [117]2 years ago
5 0

Answer:

Explanation:idk

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Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the
Inessa [10]

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

6 0
3 years ago
A car is running at a velocity of 50 miles per hour and the driver accelerates the car by 10 miles per hour square.How far the c
krek1111 [17]
It will be 80 miles and it can be done only in 16 min
7 0
2 years ago
Read 2 more answers
What happens to the force between two charges if one of the charges are doubled?
Julli [10]

Answer:

The new force between the charges becomes double of the initial force.

Explanation:

The force acting between charge particles is given by :

F=k\dfrac{q_1q_2}{r^2}

k is electrostatic constant

r is distance between charges

If one of the charges are doubled, then, q₁ = 2q₁

The new force becomes,

F'=\dfrac{2kq_1q_2}{r^2}\\\\F'=2F

So, the new force between the charges becomes double of the initial force.

7 0
3 years ago
BRAINLEST FOR CORRECT ANSWER
xxMikexx [17]

Answer:

3kg sledgehammer swung at 1.5 m/s

Explanation:

Small Sledgehammer:

Mass:3.0

Velocity:1.5

MASS×VELOCITY=MOMENTUM

3.0×1.5= 4.5 (momentum)

Large Sledgehammer:

Mass:4.0

Velocity:0.9

4.0×0.9=3.6 (momentum)

higher momentum is the smaller Sledgehammer.

3 0
2 years ago
The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
sattari [20]

Answer:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

                 

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = W_{y} / W

             W_{y} = W sin 46

     

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

 

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         Em_{f} = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

7 0
3 years ago
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