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stira [4]
2 years ago
15

A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel

eration of the rocket is given by ay=(2.60m/s3)t, where the +y-direction is upward.
Physics
1 answer:
Vikki [24]2 years ago
5 0

acceleration of rocket is given here as

a_y = 2.60* t

now we know that

\frac{dv}{dt} = 2.60t

now integrating both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

here since its given that rocket will accelerate for t = 10 s

so here we have

v = 1.30 * 10^2

v = 130 m/s

so after t = 10 s the speed of rocket will be 130 m/s upwards

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8090 [49]
V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )  
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At  α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t² 
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m

GOOD LUCK AND HOPE IT HELPS U
6 0
2 years ago
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva
AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0
2 years ago
A bike that is coasting down a steep hill increases its speed from 8.0 m/s to 14 m/s. The length of the hill is 55 meters. How m
masya89 [10]

t=5s

it was correct on my do-now

so I hope it was useful for you

8 0
2 years ago
Read 2 more answers
: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop
Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

  1. On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
  2. On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒Fa*sin(50)-Fb*sin(20)=0

⇒Fa*sin(50)=Fb*sin(20)

⇒Fa=2.24Fb

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950  

⇒Fa*cos(20)+Fb*cos(50)=950

⇒2.24*Fb*cos(20)+Fb(50)=950

⇒Fb*(2.24*cos(20)+cos(50))=950

⇒Fb=\frac{950}{2.24*cos(20)+cos(50)}

⇒Fb=\frac{950}{2.24*0.94+0.64}

⇒ Fb=\frac{950}{2.75}=345.5

⇒Fa=2.24*Fb

      =2.24*345.5

      =773.93

Therefore approximately, Fa=774 N and Fb=346 N

5 0
2 years ago
A solid cylinder of mass 12.0 kg and radius 0.250 m is free to rotate without friction around its central axis. If you do 75.0 J
faltersainse [42]

Answer:

20 rad/s

Explanation:

mass, m = 12 kg

radius, r = 0.250 m

Moment of inertia of cylinder, I = 1/2 mr²

I = 0.5 x 12 x 0.250 x 0.250 = 0.375 kgm^2

Work done = Change in kinetic energy

Initial K = 0

Final K = 1/2 Iω²

W = 1/2 Iω²

ω² = 2W/ I = 2 x 75 / (0.375)

ω = 20 rad/s

Thus, the final angular velocity is 20 rad/s .

8 0
2 years ago
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