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liberstina [14]
2 years ago
14

A toy rocket is launched with an initial velocity of 12.0 m/s in the horizontal direction from the roof of a 30.0-m-tall buildin

g. The rocket’s engine produces a horizontal acceleration of \left(1.60 m / s^{3}\right) t, in the same direction as the initial velocity, but in the vertical direction the acceleration is g , down-ward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?
Physics
1 answer:
OlgaM077 [116]2 years ago
7 0

Solution :

The motion in the y direction.

The time taken by the toy rocket to hit the ground,

$S=ut+\frac{1}{2}at^2$

S = distance travelled = 30 m

u = 0 m/s

a = $9.8 \ m/sec^2$

t= time in seconds

Therefore, $30 =\frac{1}{2}9.8 t^2$

t = 2.47 sec

Now motion in the x direction,

u = 12 m/sec

$a=1.6 t \ m/sec^3$

Upon integration 'v' with respect to 't'

$v=\frac{1.6t^2}{2}+12$

Once again integrating with respect to t,

$s=\frac{1.6t^3}{6}+12 t$

$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$

  = 0.0176+29.64

   = 29.65 m

Therefore, the toy rocket will hit the ground at 29.65 m from the building.

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Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

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Which three characteristics make iron a good blackbody radiator?
victus00 [196]

Answer:

A, C, and D

Explanation:

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3 years ago
A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?
Neko [114]

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

a is its acceleration

In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

a=\frac{F}{m}=\frac{125 N}{25.0 kg}=5 m/s^2

5 0
3 years ago
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lana66690 [7]

Answer:

a = (v2 - v1) / t

From A to B    (8 - 4) m/s / 1 s = 4 m / s^2

From A to D    ( 7 - 4) m/s / 5 s = .6 m / s^2

Note these equations hold for "uniform" values

They say nothing about the acceleration at intermediate points - the equation just says that his average speed increased from 4 m/s to 7 m/s during a 5 sec period

6 0
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A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him
juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal  direction

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v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}

Put the value into the formula

v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}

v_{c}=0.0345\ m/s

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

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