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Ghella [55]
1 year ago
6

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on

e test subject to black out and have temporary blindness.)
Physics
1 answer:
Wewaii [24]1 year ago
6 0

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

To learn more about Attention here:

brainly.com/question/28500124

#SPJ4

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After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg
kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

a)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s

<u>v₂ = 4.2 m/s</u>

<u></u>

b)

We will use the law of conservation of momentum here:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)

<u>v₂ = 5 m/s</u>

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3 years ago
A plane's average speed between two cities is 625 km/hr. If the trip takes 2.5 hours, how far does the plane fly?
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Answer: 1,562.5 Km

Explanation: If it takes one hour to travel 625 kilometers, then, in two and a half hours, you should travel 1,562.5 Kilometers because of 625x2.5=1562.5 kilometers

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