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Aleonysh [2.5K]
3 years ago
13

How much force does it take to accelerate a 50.8 kg person at 3.50 m/s^2?

Physics
1 answer:
Vikki [24]3 years ago
8 0

Apply Newton's second law to the person's motion:

F = ma

F = net force, m = mass, a = acceleration

Given values:

m = 50.8kg, a = 3.50m/s²

Plug in and solve for F:

F = 50.8(3.50)

F = 178N

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The fastest pitched baseball was clocked at 46 m/s. assume that the pitcher exerted his force (assumed to be horizontal and cons
Zielflug [23.3K]
Using the Equation:
                                 v² = vi² + 2 · a · s    → Eq.1
where,
v = final velocity 
vi = initial velocity 
a = acceleration 
s = distance 

<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,

Solving Eq.1 for acceleration,
 
</span></span> v² = vi² + 2 · a · s
 v² = 0 + 2 · a · s
 v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span> 
a = 1058 m/s</span>² 

<span>Now applying Newton's 2nd law of motion,
 </span>
<span>F = ma
   = 0.145</span>×<span>1058

F = 153.4 N</span>
8 0
3 years ago
Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s
Ksivusya [100]

The situation (heat going through the ceiling) describes
conduction ...
heat going from one place to another by
soaking through some material.

A).  This is the one.  Heat goes from from the marshmallow
to your hand by soaking through the wire.   This is conduction too.

B).  No.  The heat in the room goes from the floor to the ceiling
because the warm air rises and carries it there.  This is convection.

C).  No.  There's nothing for the heat to soak through between
the sun and the roof, and nothing that can move from the sun
to the roof and bring the heat with it.  This is radiation.

D).  No.  Cold water sinks from the surface to the bottom because
warm water rose from the bottom to the surface, taking heat with it. 
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5 0
3 years ago
Read 2 more answers
What is the approximate amount of thrust you need to apply to the lander to keep it's velocity roughly constant
earnstyle [38]

The approximate amount of thrust(force) you need to apply to the lander to

keep its velocity roughly constant is zero.

<h3>What is Newton's second law of motion?</h3>

Newton's second law of motion states that the acceleration the force acting

on the object is directly proportional to its rate of change of momentum.

F = m a

If the object is moving with uniform velocity, it simply means that the

acceleration is zero, and  the corresponding force will also be zero.  

Read more about Constant velocity here brainly.com/question/3052539

4 0
2 years ago
In 1986, several robotic spacecrafts were sent into space to study the Halley's Comet. Which of these statements best explains w
maks197457 [2]
No spacecraft has been built yet that was able to absorb harmful
radiations in space, change weather conditions on Earth, or destroy
meteors and comets which might strike Earth.

We should continue to send robotic spacecrafts into space
because they help discard some myths about objects in space. 
In other words, they help us learn things that we never knew before.

4 0
3 years ago
A rope of length L has circular cross-sectional area A and density rho = m/V , where m is the mass of the rope and V = A · L is
hram777 [196]

Answer: µ = ρ¹ * A¹

Where x=1 and y=1

Explanation: According to the question, the mass per unit length (µ) is related to the density (ρ) and area A are related by the formulae below

µ = ρ * A

The dimension for each of these quantities is given below

Since µ is mass per unit length, unit is Kg/m and the dimension is ML^-1

ρ is density with unit kg/m³ and the dimension is ML^3

A is area with unit m², thus the dimension is M^2

Note that using dimensional analysis means we will be using the 3 fundamental quantities (mass, length and time) in our analysis.

Their dimensions below

Mass = M

Length = L

Time = T

Since the mass per unit length is related to density and area, we have a mathematical equation to provide a solution as shown below

µ = ρ^x * A^y.

By getting the power of x and y we will be able to get the formula that relates the quantities.

This is done by slotting in the dimensions of the respective quantities.

ML^-1 = (ML^-3)^x * (L²) ^y

By using law of indices on the right hand side of the equation, we have that

ML^-1 = (M^x * L^-3x) * (L^2y)

Also applying law of indices on the right hand side, we have that

ML^-1 = (M^x) * (L^-3x +2y)

The next step is to relate equal variables on both sides

For the M variable

M¹ = M^x which results to

x = 1

For the L variable

L^-1 = L^-3x+2y which results to

-1 = - 3x +2y

But x = 1

We have that

-1 = - 3(1) +2y

-1 = - 3 + 2y

-1 +3= 2y

2 = 2y

y = 1

Thus x=1 and y=1 and the formulae that relates the quantities is

µ = ρ¹ * A¹

3 0
3 years ago
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