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Hunter-Best [27]

3 years ago

13

Sarah is hanging a 1.5 -kg picture on the wall. She holds it in place with a horizontal force while she checks to see if it is p

ositioned correctly. The coefficient of static friction between the picture and the wall is 0.6. Find the minimum force that stash needs to apply in order to prevent the picture form falling?

1 answer:

NNADVOKAT [17]3 years ago

4 0

Answer:

24.5 N

Explanation:

Draw a free body diagram of the picture. There are four forces:

Weight force mg pulling down.

Friction force Nμ pushing up.

Normal force N pushing left.

Applied force F pushing right.

Sum of the forces in the y direction:

∑F = ma

Nμ − mg = 0

N = mg / μ

Sum of the forces in the x direction:

∑F = ma

F − N = 0

F = N

Substitute and solve:

F = mg / μ

F = (1.5 kg) (9.8 m/s²) / (0.6)

F = 24.5 N

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Explanation:

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3 years ago

The record height of a man to date is 8 feet 11 inches (107 inches). If all men had identical body types, their weights would

Alisiya [41]

Answer:

679.08 lbs

Explanation:

Here it is given that

$m\propto h^3$

$m_1$ = Mass of first person = 175 pounds

$h_1$ = Height of first person = 70 inches

$m_2$ = Mass of second person

$h_2$ = Height of second person = 110 inches

$\frac{m_1}{m_2}=\frac{h_1^3}{h_2^3}\\\Rightarrow m_2=\frac{m_1h_2^3}{h_1^3}\\\Rightarrow m_2=\frac{175\times 110^3}{70^3}\\\Rightarrow m_2=679.08\ lb$

The weight of the second person would be 679.08 lbs

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4 years ago

if a runner's is 400 w as she runs,how much chemical energy does she convert into other forms in 10.0 minutes?

Fiesta28 [93]

Explanation:

The runner runs for a total time of

\Delta t=10.0 min= 600 sΔt=10.0min=600s

The energy converted by the runner during this time is equal to the power of the runner times the total time:

E=P \Delta t=400 W \cdot 600 s =2.4 \cdot 10^5 JE=PΔt=400W⋅600s=2.4⋅10

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7 0

4 years ago

10. A triply ionized beryllium atom is in the ground state. It absorbs energy and makes a transition to the n = 5 excited state.

Xelga [282]

Answer:

$\lambda=1282\ nm$

Explanation:

$E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

For transitions:

$Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

$\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J$

Also, $\Delta E=\frac {h\times c}{\lambda}$

Where,

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

So,

$\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J$

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

So,

$\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m$

Given, $n_i=5\ and\ n_f=3$

$\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (\frac{1}{5^2} - \dfrac{1}{3^2})}\ m$

$\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179\left(|\frac{1}{25}-\frac{1}{9}\right)|}\ m$

$\lambda=\frac{19.878}{10^8\times \:2.179\left(|-\frac{16}{225}\right|)}\ m$

$\lambda= 1.2828\times10^{-6}$

1 m = 10⁻⁹ nm

$\lambda=1282\ nm$

6 0

3 years ago

An electron is moving at a constant speed of 61 m/s on a circle of radius 3.8 m.

alexira [117]

Charge on electron is 1.6x10^-19 Coulombs.In uniform circular motion v = r omega. omega = v/r. omega= 61/3.8 radians per second - cycle is 2pi radians. (61/3.8)x2pi cycles per second. A charge circulation rate of (61/3.8)x2pix1.6x10^-19 Coulombs/second, or Amps. Which is a pretty small value of a current. Usually there many more electrons moving much faster, I think.

7 0

3 years ago