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Sav [38]
2 years ago
15

A space rocket is launched and accelerates uniformly to 160 m/s in 4.5 s. Calculate the acceleration of the rocket.​

Physics
1 answer:
-BARSIC- [3]2 years ago
6 0

Answer:

<h2>35.56 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

v is the velocity

t is the time

From the question we have

a =  \frac{160}{4.5}  = 35.55555...

We have the final answer as

<h3>35.56 m/s²</h3>

Hope this helps you

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4.) An apartment building is on fire and a guy is trapped on the fire escape ladder. There is a
Tems11 [23]

Answer:

5.3 m/s

Explanation:

First, find the time it takes for him to fall 7m.

y = y₀ + v₀ t + ½ at²

0 = 7 + (0) t + ½ (-9.8) t²

0 = 7 − 4.9 t²

t ≈ 1.20 s

Now find the velocity he needs to travel 6.3m in that time.

x = x₀ + v₀ t + ½ at²

6.3 = 0 + v₀ (1.20) + ½ (0) (1.20)²

v₀ ≈ 5.27 m/s

Rounded to two significant figures, the man must run with a speed of 5.3 m/s.

3 0
3 years ago
Can you please help me
konstantin123 [22]

Answer:

I dont know sorryyyyyyy

4 0
3 years ago
A meter stick is suspended vertically at a pivot point 22 cm from the top end. It is rotated on the pivot until it is horizontal
suter [353]

Answer:

5.82812 rad/s

Explanation:

L = Length of meter stick = 1 m = 100 cm

m_c = The center of mass of the stick = \frac{L}{2}-0.22=0.5-0.22=0.28\ m

\omega = Angular velocity

Moment of inertia of the system is given by

I=I_c+mr^2\\\Rightarrow I=\frac{mL^2}{12}+mr^2\\\Rightarrow I=\frac{m1^2}{12}+m0.28^2\\\Rightarrow I=m(\frac{1}{12}+0.0784)

As the energy in the system is conserved

mgh=I\frac{\omega^2}{2}\\\Rightarrow mgh=m(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow gh=(\frac{1}{12}+0.0784)\frac{\omega^2}{2}\\\Rightarrow \omega=\sqrt{\frac{2gh}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=\sqrt{\frac{2\times 9.81\times 0.28}{\frac{1}{12}+0.0784}}\\\Rightarrow \omega=5.82812\ rad/s

The maximum angular velocity is 5.82812 rad/s

4 0
3 years ago
Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
2 years ago
A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The
USPshnik [31]

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

4 0
3 years ago
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