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3 years ago

A 65.0 kg skier slides down a 37.20 slope with mu = 0.107.What is the friction force?

Physics

1 answer:

zmey [24]3 years ago

3 0

Answer:

54.3N

Explanation:

The normal force is perpendicular to the slope, so:

Normal Force = cos(37.2)(9.8*65).......507.39N

F(friction)=mu*F(normal)

F(friction)=(0.107)(507.39)

F(friction)=54.3N

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A penny is dropped from the top of a tower. It hits the ground below after 2.5 s. How tall was the tower?

pychu [463]

Answer:

C. 30.6m

Explanation:

To find the height of the tower, we are to use Newtons law of motion to solve this problem. Since the penny is falling from the top of the tower, it is acted by the acceleration due to gravity. The formula to be used is:

$H=ut+\frac{1}{2}gt^2$

Where H is the height of the tower, t is the time taken to hit the ground, u is the initial velocity and g is the acceleration due to gravity.

Given that, t = 2.5 s, g =9.8 m/s², u = 0 m/s (at the top of tower)

$H=ut+\frac{1}{2}gt^2\\\\H=0(2.5)+ \frac{1}{2}(9.8)(2.5)^2\\\\H=30.6\ m$

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3 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth = 5.98 × 10²⁴ kg

Radius of Earth = $6.38\times 10^6\ m$

The gravitational force will balance the centripetal force

$\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}$

$T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$

From Kepler's law we have relation

$T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m$

Distance from the center of the Earth would be

$42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}$

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3 years ago

When an object is fully magnetized, all of its magnetic domains will be ?

kkurt [141]

The answer is to this question D

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4 years ago

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Uranium-235 decays to thorium-231 with a half-life of 700 million years. When a rock was formed, it contained 6400 million urani

Dahasolnce [82]

Answer:

proof in explanation

Explanation:

First, we will calculate the number of half-lives:

$n = \frac{t}{t_{1/2}}$

where,

n = no. of half-lives = ?

t = total time passed = 2100 million years

$t_{1/2}$ = half-life = 700 million years

Therefore,

$n = \frac{2100\ million\ years}{700\ million\ years}\\\\n = 3$

Now, we will calculate the number of uranium nuclei left ( $n_u$ ):

$n_u = \frac{1}{2^{n} }(total\ nuclei)\\\\n_u = \frac{1}{2^{3} }(6400\ million)\\\\n_u = \frac{1}{8}(6400\ million)\\\\n_u = 800\ million$

and the rest of the uranium nuclei will become thorium nuclei ( $u_{th}$ )

$n_{th} = total\ nuclei - n_u\\n_{th} = 6400\ million-800\ million\\n_{th} = 5600\ million$

dividing both:

$\frac{n_{th}}{n_u}=\frac{5600\ million}{800\ million} \\\\n_{th} = 7n_u$

<u>Hence, it is proven that after 2100 million years there are seven times more thorium nuclei than uranium nuclei in the rock.</u>

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3 years ago

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Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.

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