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2 years ago

A 38 kg crate rests on a floor. A horizontal pulling force of 170 N is needed to start the crate

Physics

1 answer:

Mandarinka [93]2 years ago

5 0

Answer:

0.456033049

Explanation:

$F=\mu N$ where N=mg hence

$F=\mu mg$ where m is mass of object, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ , $\mu$ is the coefficient of static friction and F is the applied force.

Making $\mu$ the subject we obtain

$\mu=\frac {mg}{N}$ and substituting m for 38 Kg, g for $9.81 m/s^{2}$ and 170 N for F we obtain

$\mu=\frac{170} {38*9.81}=0.456033049$

Therefore, the coefficient of static friction is 0.456033049

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Sheila (m=56.8 kg) is in her saucer sled moving at 12.6 m/s at the bottom of the sledding hill near Bluebird Lake. She approache

FromTheMoon [43]

Answer:

y = 54.9 m

Explanation:

For this exercise we can use the relationship between the work of the friction force and mechanical energy.

Let's look for work

W = -fr d

The negative sign is because Lafourcade rubs always opposes the movement

On the inclined part, of Newton's second law

Y Axis

N - W cos θ = 0

The equation for the force of friction is

fr = μ N

fr = μ mg cos θ

We replace at work

W = - μ m g cos θ d

Mechanical energy in the lower part of the embankment

Em₀ = K = ½ m v²

The mechanical energy in the highest part, where it stopped

$Em_{f}$ = U = m g y

W = ΔEm = $Em_{f}$ - Em₀

- μ m g d cos θ = m g y - ½ m v²

Distance d and height (y) are related by trigonometry

sin θ = y / d

y = d sin θ

- μ m g d cos θ = m g d sin θ - ½ m v²

We calculate the distance traveled

d (g syn θ + μ g cos θ) = ½ v²

d = v²/2 g (sintea + myy cos tee)

d = 9.8 12.6 2/2 9.8 (sin16 + 0.128 cos 16)

d = 1555.85 /7.8145

d = 199.1 m

Let's use trigonometry to find the height

sin 16 = y / d

y = d sin 16

y = 199.1 sin 16

y = 54.9 m

8 0

2 years ago

What mass needs to be attached to a spring with a force constant of 7N/m in order to make a simple harmonic oscillator oscillate

Alex_Xolod [135]

Answer:

Mass will be 4.437 kg

Explanation:

We have given force constant k = 7 N/m

Time period of oscillation T = 5 sec

So angular frequency $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{5}=1.256rad/sec$

We know that angular frequency is given by

$\omega =\sqrt{\frac{k}{m}}$

$1.256 =\sqrt{\frac{7}{m}}$

Squaring both side

$1.577 =\frac{7}{m}$

m = 4.437 kg

6 0

2 years ago

A force of 6 N is used to open a door, wherein the distance of the force to the axis of rotation is 80 cm. If the angular accele

Luda [366]

Explanation:

∑τ = Iα

(6 N) (0.80 m) = I (0.5 rad/s²)

I = 9.6 kg m²

7 0

2 years ago

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie

Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0

2 years ago

Read 2 more answers

A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m

3 0

2 years ago