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2 years ago

5

A coin placed on the cover of a book just begins to move when the cover makes an angle of 38 degrees with the horizontal. What i

s the coefficient of static friction?

1 answer:

RSB [31]2 years ago

6 0

Answer:

m g sin theta = force of object along incline due to gravity

N μ = frictional of incline on object where N is the normal force

N = m g cos theta force perpendicular to incline

m g sin theta = N μ = μ m g cos theta

μ = tan theta = tan 38 = .78

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A student is investigating thermal energy transfer. The student touches a piece of metal to a hot object, causing the metal to h

aliina [53]

It should be C. Conduction

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3 years ago

shutvik [7]

Answer:

a) P = 149140[w]; b) 1491400[J]; c) v = 63.06[m/s]

Explanation:

As the solution to the problem indicates, we must convert the power unit from horsepower to kilowatts.

P = 200 [hp]

$200[hp] * 745.7 [\frac{watt}{1 hp}]\\149140[watt]$

Now the power definition is known as the amount of work done in a given time

P = w / t

where:

w = work [J]

t = time [s]

We have the time, and the power therefore we can calculate the work done.

w = P * t

w = 149140 * 10 = 1491400 [J]

And finally, we can calculate the velocity using, the expression for kinetic energy

$E_{k}=w=0.5*m*v^{2}\\ where:\\v = velocity[m/s]\\m=mass=750[kg]\\w=work=1491400[J]\\$

The key to solving this problem is to recognize that work equals kinetic energy

$v=\sqrt{\frac{w}{0.5*m}} \\v=\sqrt{\frac{1491400}{0.5*750}} \\v=63.06[m/s]$

3 0

3 years ago

A body can have zero average velocity but not zero average speed. Why?

bogdanovich [222]

Displacement can be zero so velocity is zero but the distance won’t be zero so speed won’t be zero

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3 years ago

A conductor is a material in which what can move through easily? Electrons Water Condensation Oil

kondaur [170]

Answer:

Electrons

Explanation:

:) hope this helps. Have a great day!

4 0

3 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity $R_{0}=10\ mCi$

Time $t_{1}=4\ hours$

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

$R=R_{0}e^{-\lambda t}$

$\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})$

Put the value into the formula

$\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})$

$\lambda=0.0000154\ s^{-1}$

$\lambda=1.55\times10^{-5}\ s^{-1}$

We need to calculate the half life

Using formula of half life

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}$

Put the value into the formula

$T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}$

$T_{\dfrac{1}{2}}=44.719\times10^{3}\ s$

$T_{\dfrac{1}{2}}=11.3\ hr$

(b). We need to calculate the value of N₀

Using formula of $N_{0}$

$N_{0}=\dfrac{3.70\times10^{6}}{\lambda}$

Put the value into the formula

$N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}$

$N_{0}=2.38\times10^{11}\ nuclei$

(c). We need to calculate the sample's activity

Using formula of activity

$R=R_{0}e^{-\lambda\times t}$

Put the value intyo the formula

$R=10e^{-(1.55\times10^{-5}\times30\times3600)}$

$R=1.87\ mCi$

Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

The half life is 11.3 hr.

(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

(c). The sample's activity is 1.87 mCi.

4 0

3 years ago