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Artemon [7]
3 years ago
7

A 590-kg rocket is at rest on the launch pad. what upward thrust force is needed to accelerate the rocket uniformly to an upward

speed of 28 m/s in 3.3 s?
Physics
2 answers:
fgiga [73]3 years ago
5 0

<u>Answer:</u> The upward thrust force needed is 5003.2 N

<u>Explanation:</u>

Acceleration is defined as the rate of change of velocity with respect to time.

Mathematically,

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

where,

v = final velocity = 28 m/s

u = initial velocity = 0 m/s

t = time taken = 3.3 s

Putting values in above equation, we get:

a=\frac{28-0}{3.3}=8.48m/s^2

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=m\times a

where,

F = force exerted on the rocket = ?

m = mass of the rocket = 590 kg

a = acceleration of the rocket = 8.48m/s^2

Putting values in above equation, we get:

F=590kg\times 8.48m/s^2\\\\F=5003.2N

Hence, the upward thrust force needed is 5003.2 N

BARSIC [14]3 years ago
3 0
V(final)=V₀+a(t), V(final)=28, V₀=0, t=3.3, solve for a. a=28/3.3=8.5m/s^2. F=ma, F=?, m=590kg, a=8.5m/s^2, F=5015N. 
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A 5kg block is pulled across a table by a horizontal force of 40 N with a frictional force of 8 N opposing the motion. Draw a fo
Taya2010 [7]

Answer:

Net force on the block is 32 N.

Acceleration of the object is 6.4 m/s².

Explanation:

Let the acceleration of the object be a m/s².

Given:

Mass of the block is, m=5\ kg

Force of pull is, F=40\ N

Frictional force on the block is, f=8\ N

The free body diagram of the object is shown below.

From the figure, the net force in the forward direction is given as:

F_{net}=F-f=40-8=32\ N

Now, from Newton's second law of motion, net force is equal to the product of mass and acceleration. So,

F_{net}=ma\\32=5a\\a=\frac{32}{5}=6.4\ m/s^2

Therefore, the acceleration of the object in the forward direction is 6.4 m/s².

3 0
3 years ago
The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density
OLga [1]

Answer:

V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s

Explanation:

The volume flow rate of the blood in the artery can be given by the following formula:

V = Av

where,

V = Volume flow rate = ?

A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²

v = velcoity = 0.28 m/s

Therefore,

V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)

<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>

4 0
3 years ago
The cloud of interstellar dust and gas that forms a star is known as a
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A car is traveling with a velocity of 24.4 m/s. It accelerates at a constant rate of 3.2m/s2. If the acceleration lasts for 5.6
larisa86 [58]

Answer: <em>4</em><em>2</em><em>.</em><em>3</em><em>2</em><em> </em><em>ms-1</em>

Explanation:

v = u+ at

v = 24.4 + ( 3.2×5.6)

v = 42.32 ms-1

6 0
3 years ago
A 75 N box of oranges is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each
shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Data:

a= -0.9  m/s²,  

g = 9.81 m/s² : acceleration due to gravity

W= 75 N :  Block weight

W= m*g  

m =  W/g = 75/9.8= 7.65 kg :  Block mass

Friction force : Ff

Ff= μk*N

μk: coefficient of kinetic friction

N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax    

20-Ff= m*ax    

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

μk = 0.26885

4 0
3 years ago
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