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Nookie1986 [14]

3 years ago

15

A force of 36.0 N is required to start a 3.0-kg box moving across a horizontal concrete floor. Part A) What is the coefficient o

f static friction between the box and the floor? Express your answer using two significant figures. Part B) If the 36.0-N force continues, the box accelerates at 0.54 m/s^2. What is the coefficient of kinetic friction? Express your answer using two significant figures.

1 answer:

matrenka [14]3 years ago

4 0

Answer:

A) 1.2

B) 1.1

Explanation:

A)

F = force required to start the box moving = 36.0 N

m = mass of the box = 3 kg

$F_{g}$ = weight of the box = mg = 3 x 9.8 = 29.4 N

$F_{n}$ = normal force acting on the box by the floor

normal force acting on the box by the floor is given as

$F_{n}$ = $F_{g}$ = 29.4

$F_{s}$ = Static frictional force = F = 36.0 N

$\mu _{s}$ = Coefficient of static friction

Static frictional force is given as

$F_{s}$ = $\mu _{s}$ $F_{n}$

36.0 = $\mu _{s}$ (29.4)

$\mu _{s}$ = 1.2

B)

a = acceleration of the box = 0.54 m/s²

F = force applied = 36.0 N

$f_{k}$ = kinetic frictional force

$\mu _{k}$ = Coefficient of kinetic friction

force equation for the motion of the box is given as

F - $f_{k}$ = ma

36.0 - $f_{k}$ = (3) (0.54)

$f_{k}$ = 34.38 N

Coefficient of kinetic friction is given as

$\mu _{k}=\frac{f_{k}}{F_{n}}$

$\mu _{k}=\frac{34.38}{29.4}$

$\mu _{k}$ = 1.1

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Sunny_sXe [5.5K]

Answer:

The speed of the car at the apex of the loop must be grater than 2.45 m/s

Explanation:

In order for the car to not fall off the track at the apex of the loop, the norm force of the track at the apex must be greater than zero.

Assuming frictionless life on the track which is also to have a perfectly circular shape near the top (radius being 0.25m), the norm force of the track and gravity both point down and result in the centripetal force:

$F_c=F_N+F_g$

The formula for centripetal force on a circular trajectory is

$F_c = m\frac{v^2}{r}$

and so the condition for the car to stay on the track can be written as

$m\frac{v^2}{r} = F_N + mg\implies F_N = m\frac{v^2}{r}-mg>0\\\implies |v| >\sqrt{gr}=\sqrt{9.8\frac{m}{s^2}0.25m}=2.45\frac{m}{s}$

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3 years ago

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neonofarm [45]

Answer:

The impulse exerted by the ball is 7.2 kg.m/s

The average force exerted on the ball by the bat is 600 N.

Explanation:

Given;

mass of the baseball, m = 0.144 kg

velocity of the baseball, v₁ = 20 m/s

velocity of the batter, v₂ = -30 m/s (opposite direction to the ball's speed)

The impulse exerted by the ball is calculated as follows;

J = ΔP = mv₁ - mv₂

ΔP = m(v₁ - v₂)

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ΔP = 0.144 ( 20 + 30 )

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The average force exerted on the ball by the bat is calculated as;

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3 years ago

Svet_ta [14]

Answer:

See below ~

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An object will sink in water when its density is greater than that of water, which is 1 g/cm³.

Volume of the box is <u>1331 cm³</u>. (11³)

Maximum mass of sand will be 1331 g. [because 1331/1331 = 1 g/cm³]

Volume of sand = Mass of sand / Density of sand
Volume (sand) = 1331/3.5
Volume (sand) = 380.29 cm³

If the volume of sand is <u>greater than 380.29 cm³</u>, the box will sink in water.

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3 years ago

For the population, scores on the test are normally distributed with μ = 70 and σ = 15. The sample of n = 25 students had a mean

AysviL [449]

Answer:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

$p_v =2*P(t_{24}>1.67)=0.108$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

Explanation:

1) Data given and notation

$\bar X=75$ represent the sample mean

$\sigma=15$ represent the standard deviation for the population

$n=25$ sample size

$\mu_o =70$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is different from 70, the system of hypothesis would be:

Null hypothesis: $\mu = 70$

Alternative hypothesis: $\mu \neq 70$

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{75-70}{\frac{15}{\sqrt{25}}}=1.67$

Calculate the P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=25-1=24$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{24}>1.67)=0.108$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not significant different from 70.

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4 years ago