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iris [78.8K]
3 years ago
9

While riding in a hot air balloon, which is steadily decreasing at a speed of 1.14 m/s, you accidentally drop your cell phone. H

ow far (in m) is the cell phone below the balloon after this time?
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0

Answer:

The distance covered by the balloon is 47.52 meters.

Explanation:

Given that,

Initial speed of the balloon, u = 1.14 m/

Let us assumed we need to find the distance covered by the balloon after t = 3 second. Let d is the distance covered by the balloon. It can be given by :

d=ut+\dfrac{1}{2}at^2

Here, a = g

d=ut+\dfrac{1}{2}gt^2

d=1.14\times 3+\dfrac{1}{2}\times 9.8\times (3)^2

d = 47.52 meters

So, the distance covered by the balloon is 47.52 meters. Hence, this is the required solution.

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During a thunderstorm, an observer notes that 10 s elapsed between the lightning flash and the sound of the thunder. what is the
Masja [62]
The answer is A. every sec means a mile.hope this helped
3 0
3 years ago
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially
puteri [66]

Answer:

Vb = v/2

Half of the speed of Car A before the collision

Explanation:

This problem can be solved by using conservation of momentum.

m_{A}*v_{A1} + m_{B}*v_{B1} = m_{A}*v_{A2} + m_{B}*v_{B2}

Since car B was stopped before the collision and its mass is twice the mass of A:

v_{B1} = 0\\m_{B} = 2*m_{A}

Also, since it is an elastic collision, car A stops after hitting car B. Rewriting the equation:

m_{A}*v_{A1} + 2m_{A}*0 = m_{A}*0 + 2m_{A}*v_{B2}\\m_{A}*v_{A1} = 2m_{A}*v_{B2}\\v_{B2}=\frac{v_{A1}}{2}

Therefore, the speed of car B after the collision is half of the speed of car A before the collision (v)

Vb = v/2

8 0
4 years ago
What horizontal speed must a pumpkin be thrown to hit a car 13.4 meters away from a building which stands 10.4 meters tall?
Ilia_Sergeevich [38]

Answer:

V₀ₓ = 9.2 m/s

Nearest answer:

D) 8.9 m/s

Explanation:

First we find the time taken by the pumpkin to hit the car. For that purpose we apply 2nd equation of motion to the pumpkin:

h = V₀y t + (1/2)gt²

where,

h = height of building = 10.4 m

V₀y = vertical component of initial speed = 0 m/s

t = time = ?

g = 9.8 m/s²

Therefore,

10.4 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (10.4 m)(2)/(9.8 m/s²)

t = √[2.122 s²]

t = 1.45 s

Now, we analyze horizontal motion for horizontal component of initial velocity. We assume air friction to be zero so that the horizontal motion is uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance between building and car = 13.4 m

V₀ₓ = Horizontal Component of Initial Velocity = ?

Therefore,

13.4 m = V₀ₓ(1.45 s)

V₀ₓ = 13.4 m/1.45 s

<u>V₀ₓ = 9.2 m/s</u>

6 0
3 years ago
A 5 inch tall balloon shoot doubles in height every 3 days. if the equation y=ab^x, where is x is the number of doubling periods
VashaNatasha [74]
In this item if we let x be the number of doubling period, that is number of days divided by 3, and y be the height of the bamboo shoot, in the equation
                                     y  = ab^x
The value of a is 5 because that is the original height of the shoot and b is 2 because we are talking about the doubling rate. The equation is therefore,
                                    y = 5(2^x)
3 0
3 years ago
A 0.4-kg cart, traveling on a horizontal air track with a speed of 6 m/s, collides with a stationary 0.8-kg cart. The carts stic
steposvetlana [31]

Answer:

B.1.6 N*s

Explanation:

According to the principle of conservation of momentum, we have:

\Delta p=0\\p_f-p_i=0\\p_f=p_i\\v_fm_f=m_1v_1+m_2v_2

The final mass is obtained adding the masses of the two cars since they stick together after the collision. So, m_f=m_1+m_2. Recall that the 0.4 kg cart collides with the stationary 0.8-kg cart. So v_2=0

v_f(m_1+m_2)=m_1v_1\\v_f=\frac{m_1v_1}{m_1+m_2}\\v_f=\frac{0.4kg(6\frac{m}{s})}{0.4kg+0.8kg}\\v_f=2\frac{m}{s}

The magnitude of the impulse is defined as the mass multiplied by the change in the speed:

I=m\Delta v \\I=m(v_f-v_i)\\I=0.8kg(2\frac{m}{s}-0)\\I=1.6N\cdot s

3 0
4 years ago
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