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3 years ago

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You are pulling your little sister on her sled across an icy (frictionless) surface. When you exert a constant horizontal force

of 120 N , the sled has an acceleration of 2.3 m/s2 . Part A If the sled has a mass of 8.0 kg , what is the mass of your little sister

1 answer:

Tpy6a [65]3 years ago

4 0

Answer:

Mass of Little Sister = 44.17 kg

Explanation:

From Newton's second law of motion, the magnitude of force applied on the sled is given by the following formula:

F = ma

where,

F = Force Applied = 120 N

a = Acceleration = 2.3 m/s²

m = Mass of Sled + Mass of Little Sister = 8 kg + Mass of Little Sister

Therefore,

120 N = (2.3 m/s²)(8 kg + Mass of Little Sister)

(120 N)/(2.3 m/s²) = 8 kg + Mass of Little Sister

Mass of Little Sister = 52.17 kg - 8 kg

<u>Mass of Little Sister = 44.17 kg</u>

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A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .

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2 years ago

Find a first-degree polynomial function p1 whose value and slope agree with the value and slope of f at x = c. f(x) = cot(x), c

Neporo4naja [7]

The correct answer is y=-2x+(1/2)

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Y = -2x + C

1 = -2x π/4 + C

=) C = I + π/2

y=-2x+(1/2) is the first-degree polynomial.

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1 year ago

Which oftthe following is the least reliable source of background information for a scientific project? General internet site, g

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General internet site

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3 years ago

The displacement volume of an automobile engine is 167 in3. what is the displacement volume in liters?

forsale [732]

The displacement volume in liters is 2.74 liters.

<h3>What is displacement volume?</h3>

Displacement volume is the quantity of solvent that will be displaced by a specified quantity of a solid during dissolution.

It can also be defined as the volume displaced by the piston as it moves between top dead center and bottom dead center in a car engine.

<h3>Displacement volume in liters</h3>

1 liter = 61.02 in³

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= 167/61.02

= 2.74 liters

Thus, the displacement volume in liters is 2.74 liters.

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1 year ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

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Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

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3 years ago