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Morgarella [4.7K]
2 years ago
12

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. it accelerates upward at 30.0 m/s2 for 30.0s, then

runs out of fuel. ignore any air resistance effects. part a what is the rocket's maximum altitude?
Physics
1 answer:
AlladinOne [14]2 years ago
7 0
There are two stages to the flight: acceleration stage and deceleration stage.

m₁ = 200 kg, mass of the rocket
m₂ = 100 kg, mass of fuel
a₁ = 30.0 m/s², upward acceleration when burning fuel
Ignore air resistance and assume g = 9.8 m/s².

Acceleration stage:
The rocket starts from rest, therefore the initial vertical velocity is zero.
The distance traveled is given by
s₁ = (1/2)*(30.0 m/s²)*(30.0 s)² = 13500 m

Deceleration stage (due to gravity):
The initial velocity is u = (30.0 m/s²)*(30 s) = 900 m/s
The initial height is 13500 m
At maximum height, the vertical velocity is zero.
Let s₂ =  the extra height traveled. Then
(900 m/s)² - 2*(9.8 m/s²)*(s₂ m) = 0
s₂ = 900²/19.6 = 41326.5 m

The maximum altitude is 
s₁+s₂ = 54826.5 m

Answer: 54,826.5 m
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Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

so its weight in air will be given as

W = mg = 1.134* 9.8 = 11.11 N

now the buoyant force on the lead is given by

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

now as we know that

F_B = \rho V g

0.11 = 1000* V * 9.8

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water


#2

buoyant force on the lead block is balancing the weight of it

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight =  11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury


#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0
2 years ago
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How lon
frosja888 [35]

Answer:

t=0.704s

Explanation:

A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?

from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

V^{2} =u^{2} +2as

Vy^2=0+2*9.8*1.73sin40.5

Vy^2=22.021

Vy=4.69m/s

Vx^2=u^2+2*9.81*cos40.5

Vy^2=25.81

Vy=5.08m/s

V=(Vy^2+Vx^2)^0.5

V=47.71^0.5

V=6.9m/s

from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.

f=force applied

v=velocity final

u=initial velocity

m=mass of the toy, 0.046

f=ma

f=m(v-u)/t

v=u+at

6.9=0+9.8t

t=6.9/9.81

t=0.704s

4 0
2 years ago
The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth.
Talja [164]

Answer:

F_g=461lb_f

Explanation:

First calculate the mass of the asteroid. To do so, you need to find the volume and know the density of iron.

If r = d/2 = 645ft, then:

V = \frac{4}{3} \pi r^3

V = 1.124*10^{9}ft^3

So

\delta_{iron}=m/V=491lb/ft^3

m=V*\delta=5.519*10^{11}lb

Once you  know both masses, you can calculate the force using Newton's universal law of gravitation:

F_g=G\frac{m_1m_2}{d^2}

Where G is the gravitational constant:

G= 1.068846 * 10^{-9} ft^3 lb^{-1} s^{-2}

F_g=461lb_f

4 0
2 years ago
You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera
Nina [5.8K]

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

  • Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁  = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = \frac{2*4186*5}{0.5*25}= 3348.8J/kg C≅ 3349J/kg C

4 0
2 years ago
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