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dmitriy555 [2]
3 years ago
11

A 200 kg crate is pulled along a level surface by an engine. the coefficient of friction between crate and surface is 0.4.

Physics
2 answers:
siniylev [52]3 years ago
4 0
Normal force = 200(9.81) = 1962 N
Friction force = 1962 * .4 = 784.8
power = force * distance / time = 784.8 * 5 = 3924 Watts

Watts = Joules / sec
3924 x 180 = 706,320 Joules
N76 [4]3 years ago
3 0

Answer:

a) Power = P = 3924 watts

b) Work = W = 706320 J

Explanation:

a) Mass of the block = m = 200 kg

       Velocity = V = 5 m/s

       Time = t = 180s

       Coefficient of friction = µ = 0.4

       Gravitational acceleration = g = 9.81 m/s^2

        According to newton’s second law of motion:

                              F-f = 0

        Where,

         F = Normal force  

         f = Frictional force

          so,  

                          F = f = µmg = (0.4)(200)(9.81) =  784.8 N

          We know that:

                         P = FV = (784)(5) = 3924 watts

b)      We know that:

          W = P × t = 3924 × 180 = 706320 J

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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0
1 year ago
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0
Murrr4er [49]

Complete question:

An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.

Answer:

The value of its capacitance is 1.027 x 10⁻¹² F

Explanation:

Given;

area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²

separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m

voltage of the battery, V = 18 V

The value of its capacitance is calculated as;

C = \frac{k\epsilon_0A}{d} \\\\C = \frac{(1)(8.85\times 10^{-12})(2.9 \times 10^{-4})}{2.5 \times 10^{-3}} \\\\C = 1.027 \times 10^{-12} \ F

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F

8 0
2 years ago
A piece of metal will feel colder than a piece of wood at the same temperature. Why?
alexira [117]
Metals in general, are good heat conductors 
7 0
2 years ago
A triangular plate with height 6 ft and a base of 7 ft is submerged vertically in water so that the top is 2 ft below the surfac
xenn [34]

Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

y-6=-\frac{6}{7}(x-0)

which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

x=-\frac{7}{6}y+7

with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

When solving we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

6 0
3 years ago
I need help ASAP. This is for 15 points
kramer

Answer:

Latitude :

runs: east to west

measures : distances north and south of the equator

Longitude :

runs : north to south

measures : the distance east or west of the Prime Meridian

7 0
3 years ago
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