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3 years ago

17. A 25 kg block is initially at rest on a rough, horizontal surface. A horizontal force of 75 N is required to set

Physics

1 answer:

lapo4ka [179]3 years ago

4 0

Answer:

0.30581

0.24464

Explanation:

$\mu_s$ = Coefficient of static friction

$\mu_k$ = Coefficient of kinetic friction

$F_f$ = 75 N

$F_k$ = 60 N

Normal force

$F_n=mg\\\Rightarrow F_n=25\times 9.81\\\Rightarrow F_n=245.25\ N$

Frictional force

$F_f=\mu_sF_n\\\Rightarrow \mu_s=\frac{F_f}{F_n}\\\Rightarrow \mu_s=\frac{75}{245.25}\\\Rightarrow \mu_s=0.30581$

The coefficient of static friction is 0.30581

Kinetic force

$F_k=\mu_kF_n\\\Rightarrow \mu_k=\frac{F_k}{F_n}\\\Rightarrow \mu_s=\frac{60}{245.25}\\\Rightarrow \mu_s=0.24464$

The coefficient of kinetic friction is 0.24464

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A chair of weight 70.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =

Annette [7]

Answer:

94.67 N

Explanation:

Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.

At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.

$\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}$

Substituting 70 N for W, 41 N for F and $\theta$ for 37 degrees

N=70+41sin37=94.67441595 N and rounding off to 2 decimal places

N=94.67 N

6 0

4 years ago

A solar cooker, really a concave mirror pointed at the Sun, focuses the Sun's rays 13.8 cm in front of the mirror. Part A What i

shepuryov [24]

From the concept of optics on a curvature of a spherical mirror, the proportion for which the focal length is equivalent to half the radius of curvature is fulfilled. Mathematically this is

$f = \frac{R}{2}$

Here,

f = Focal Length

R = Radius

Rearranging to find the radius we have,

$R = 2f$

Replacing with our values,

R = 2(13.8cm)

R = 27.6cm

Therefore the radius of the spherical surface from which the mirror was made is 27.6cm

6 0

3 years ago

Value of x<br>Guys plz answer <br>and get marked as brainliest :)

lakkis [162]

Answer:

80⁰

Explanation:

triangle AOB = 180

So, a = 50

triangle CAB = a+a+x=180

CAB = 50+50+X=180

CAB = X = 180-100

X=80

HOPE IT HELPS

3 0

3 years ago

Read 2 more answers

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

3 years ago

Water is circulating through a closed system of pipes in a two floor apartment. On the first floor, the water has a gauge pressu

anastassius [24]

Answer:

The value of gauge pressure at outlet = -38557.224 pascal

Explanation:

Apply Bernoulli' s Equation

$\frac{P_{1}}{9810}$ + $\frac{V_{1} ^{2}}{19.62}$ + $h_{1}$ = $\frac{P_{2}}{9810}$ + $\frac{V_{2} ^{2}}{19.62}$ + $h_{2}$ --------------(1)

Where

$P_{1}$ = Gauge pressure at inlet = 3.70105 pascal

$V_{1}$ = velocity at inlet = 2.4 $\frac{m}{sec}$

$P_{2}$ = Gauge pressure at outlet = we have to calculate

$V_{2}$ = velocity at outlet = 3.5 $\frac{m}{sec}$

$h_{2} - h_{1}$ = 3.6 m

Put all the values in equation (1) we get,

⇒ $\frac{3.70105}{9810}$ + $\frac{2.4 ^{2}}{19.62}$ = $\frac{P_{2}}{9810}$ + $\frac{3.5 ^{2}}{19.62}$ + 3.6

⇒ 0.294 = $\frac{P_{2}}{9810}$ + 0.6244 + 3.6

⇒ $\frac{P_{2}}{9810}$ = 0.294 - 0.6244 - 3.6

⇒ $\frac{P_{2}}{9810}$ = - 3.9304

⇒ $P_{2}$ = - 38557.224 pascal

This is the value of gauge pressure at outlet.

3 0

3 years ago