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3 years ago

1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force

Physics

1 answer:

Arturiano [62]3 years ago

4 0

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

$F=Fr=\mu N$ where N is the weight of object and \mu is coefficient of kinetic friction.

$F=\mu N and making [tex]\mu$ the subject

$\mu=\frac {F}{N}$

Substituting F for 8 N and N for 20 N

$\mu=\frac {8}{20}=0.4$

Therefore, coefficient of kinetic friction is 0.4

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The engine of a locomotive exerts a constant force of 8.1*10^5 N to accelerate a train to 68 km/h. Determine the time (in min) t

Bumek [7]

Answer:443.1 s

Explanation:

Given

Engine of a locomotive exerts a force of $8.1\times 10^5 N$

Mass of train $=1.9\times 10^7$

Final speed (v) $=68 km/h \approx 18.88 m/s$

F=ma

so $acceleration(a) =\frac{F}{m}=\frac{8.1\times 10^5}{1.9\times 10^7}$

$a=0.042631 m/s^2$

and acceleration is

$a=\frac{v-u}{t}$

$0.042631=\frac{18.88-0}{t}$

$t=443.089 \approx 443.1 s$

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3 years ago

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Answer:-4

Explanation:because

10protone +10

14 electrone-14

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10-14=-4

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3 years ago

A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr

dalvyx [7]

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

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3 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

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$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$