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3 years ago

5

What is your height in centimeters? what is your weight in Newton's

2 answers:

Gala2k [10]3 years ago

6 0

157.48 centimeters

1201.01984 newtons

goldfiish [28.3K]3 years ago

4 0

My height is 178 centimeters.
My weight is 700N.

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If the mass of an object is 200kg and the applied force is 2600N, calculate the Acceleration.

Aliun [14]

Answer:

<em>13 m/s²</em>

Explanation:

Mass of object = 200 Kg

Applied force = 2600 N

Acceleration = ?

Solution:

Definition:

The acceleration is rate of change of velocity of an object with respect to time.

Formula:

a = Δv/Δt

a = acceleration

Δv = change in velocity

Δt = change in time

Units:

The unit of acceleration is m.s⁻².

Acceleration can also be determine through following formula,

F = m × a

a = F/m (N = kgm/s²)

a = 2600 kgm/s² / 200 Kg

a = 13 m/s²

6 0

3 years ago

An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid

mariarad [96]

Explanation:

It is given that, An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid X and 481 km from the center of asteroid Y, along the straight line joining the centers of the asteroids. We need to find the ratio of their masses.

As they are in equilibrium, the force of gravity due to each other is same. So,

$\dfrac{Gm_xM}{r^2}=\dfrac{Gm_yM}{r^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{r_x^2}=\dfrac{m_y}{r_y^2}\\\\\dfrac{m_x}{m_y}=(\dfrac{r_x^2}{r_y^2})\\\\\dfrac{m_x}{m_y}=(\dfrac{140^2}{481^2})\\\\\dfrac{m_x}{m_y}=0.0847$

So, the ratio of masses X/Y is 0.0847

5 0

3 years ago

A 2.00 kg, frictionless block s attached to an ideal spring with force constant 550 N/m. At t = 0 the spring is neither stretche

gladu [14]

Answer:

a) A = 0.603 m , b) a = 165.8 m / s² , c) F = 331.7 N

Explanation:

For this exercise we use the law of conservation of energy

Starting point before touching the spring

Em₀ = K = ½ m v²

End Point with fully compressed spring

$Em_{f}$ = $K_{e}$ = ½ k x²

Emo = $Em_{f}$

½ m v² = ½ k x²

x = √(m / k) v

x = √ (2.00 / 550) 10.0

x = 0.603 m

This is the maximum compression corresponding to the range of motion

A = 0.603 m

b) Let's write Newton's second law at the point of maximum compression

F = m a

k x = ma

a = k / m x

a = 550 / 2.00 0.603

a = 165.8 m / s²

With direction to the right (positive)

c) The value of the elastic force, let's calculate

F = k x

F = 550 0.603

F = 331.65 N

5 0

3 years ago

Consider being at the top of cliff and throwing a book off the ledge. The book leaves at an angle of 52 degrees and a velocity o

Masja [62]

The vertical distance through which the book falls is determined as 1,048.8 m.

<h3>Height of the book fall</h3>

The vertical distance through which the book falls is calculated as follows;

h = vt + ¹/₂gt²

where;

h is height of fall
v is initial vertical velocity
g is acceleration due to gravity

h = (16 x sin52)(13.4) + (0.5)(9.8)(13.4²)

h = 1,048.8 m

Thus, the vertical distance through which the book falls is determined as 1,048.8 m.

Learn more about height of fall here: brainly.com/question/15611384

#SPJ1

3 0

2 years ago

Substances A and B, initially at different temperatures, come in contact with each other and reach thermal equilibrium. The mass

Gnesinka [82]

Answer: .2) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_A\times c_A\times (T_{final}-T_A)=-[m_B\times c_B\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_A$ = mass of A = 2x

$m_B$ = mass of B = x

$T_{final}$ = final temperature = z

$T_A$ = temperature of A

$T_2$ = temperature of B

$c_A$ = specific heat capacity of A = y

$c_B$ = specific heat capacity of B = 2y

Now put all the given values in equation (1), we get

$2x\times y\times (z-T_A)=-[x\times 2y\times (z-T_B)]$

$2z=T_B+T_A$

$z=\frac{T_B+T_A}{2}$

Therefore, the final temperature of the mixture will be exactly midway between the initial temperatures of substances A and B.

3 0

3 years ago