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2 years ago

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Which is closer to the Sun—Mercury or the Earth? A. They are the same distance from the Sun. B. The Earth is closer to the Sun t

han Mercury is. C. Mercury is closer to the Sun than the Earth is. D. The distance from the Earth to the Sun cannot be determined.

1 answer:

tankabanditka [31]2 years ago

3 0

Answer:

the answer is C

Explanation:

I know because mercury is the first planet closest to the sun

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Wo siblings are arguing over what way to pull a 10 kg wagon. Boris wants to pull it to the right. Boris puts 220 N of force on t

Mrrafil [7]

Answer:

a) Then net force acting on the wagon 30 [N] in a negative direction to the left

b)Acceleration of the wagon 3 m/s² in a negative direction to the left

Explanation:

The coordinates x and y show the positive direction in x-axis ( to the right) and negative direction to the left. In north-south direction positive y values are to the north and negative values to the south

The free-body diagram shows 4 forces acting

In y-axis:

The weight ( P = m*g = 10 [kg]*9,8 [m/s²] = 98 [N] negative

Normal reaction force Fn = P = 98 [N] positive ( surface is frictionless)

∑Fy = 0 P - Fn = 0 P = Fn = 98 [N]

In x-axis:

Boris pulling force to the right (positive ) 220 [N]

Natasha pulling force to the left ( negative) 250 [N]

∑Fx = m*a

220 [N] - 250 [N] = 10 Kg * a [N] = Kg * m /s²

-30 [kg*m/s²] = 10 * Kg * a

-30/10 = - 3 m/s² = a

Sign (-) means the direction of acceleration vector is to the left (the same direction of the movement )

Then net force acting on the wagon 30 [N] in negative direction to the left

Acceleration of the wagon 3 m/s² in negative direction to the left

5 0

2 years ago

Several springs are connected as illustrated below in (a). Knowing the individual springs stiffness k1 = 20 N/m, k2 = 30 N/m, k3

Hatshy [7]

Answer:

The equivalent stiffness of the string is 8.93 N/m.

Explanation:

Given that,

Spring stiffness is

$k_{1}=20\ N/m$

$k_{2}=30\ N/m$

$k_{3}=15\ N/m$

$k_{4}=20\ N/m$

$k_{5}=35\ N/m$

According to figure,

$k_{2}$ and $k_{3}$ is in series

We need to calculate the equivalent

Using formula for series

$\dfrac{1}{k}=\dfrac{1}{k_{2}}+\dfrac{1}{k_{3}}$

$k=\dfrac{k_{2}k_{3}}{k_{2}+k_{3}}$

Put the value into the formula

$k=\dfrac{30\times15}{30+15}$

$k=10\ N/m$

k and $k_{4}$ is in parallel

We need to calculate the k'

Using formula for parallel

$k'=k+k_{4}$

Put the value into the formula

$k'=10+20$

$k'=30\ N/m$

$k_{1}$ ,k' and $k_{5}$ is in series

We need to calculate the equivalent stiffness of the spring

Using formula for series

$k_{eq}=\dfrac{1}{k_{1}}+\dfrac{1}{k'}+\dfrac{1}{k_{5}}$

Put the value into the formula

$k_{eq}=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}$

$k_{eq}=8.93\ N/m$

Hence, The equivalent stiffness of the string is 8.93 N/m.

3 0

3 years ago

A race car reaches the finish line and the driver slows down to come to a stop. If it takes the car 10 seconds to stop at a cons

boyakko [2]

Answer:

53.64 m/s

Explanation:

Applying,

a = (v-u)/t............. Equation 1

Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

make u the subject of the equation

u = v-at............. Equation 2

From the question,

Given: a = -12 mph/s = -5.364 m/s², t = 10 seconds, v = 0 m/s (comes to stop)

Substitute these values into equation 2

u = 0-(-5.364×10)

u = 0+53.64

u = 53.64 m/s

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2 years ago

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otez555 [7]

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2) Water expands when it freezes, making it less dense than just water.

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3 years ago

Mrs. Perez added a room temperature copper cube and an aluminum cube she just removed from the freezer to a beaker of boiling wa

faust18 [17]

I think the answer is A

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3 years ago