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Doss [256]
3 years ago
5

A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki

netic friction between the box and the floor?0.11 a)0.10 b)0.13 c)0.09 d)0.11
Physics
2 answers:
earnstyle [38]3 years ago
5 0

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

Rounded to the nearest hundredth, that's    <em>0.11 </em>.      (choice 'd')


Tems11 [23]3 years ago
4 0
The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.
If the box is sliding at constant speed, and not speeding up or slowing down, <span>that means that the horizontal forces on it add up to zero.</span>
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Pete is driving down 7th Street. He drives 300 meters in 18 seconds. Assuming he does not speed up or slow down, what is his spe
Korolek [52]

Answer:

16.67m/s

Explanation:

Given parameters:

Distance Pete drove  = 300m

Time taken  = 18s

Unknown:

Speed  = ?

Solution:

Speed is the distance traveled per unit of time.

It is mathematically expressed as;

   Speed  = \frac{distance}{time}

Insert the parameters and solve;

  Speed  = \frac{300}{18}  = 16.67m/s

3 0
3 years ago
What type of disorder is schizophrenia?
Oksana_A [137]
The correct answer is A because the main disorder is schizophrenia. It includes all the previous subtypes: catatonic, disorganized, paranoid, residual, and undifferentiated. It's a psychosis, which means that what seems real to you isn't.
6 0
3 years ago
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
4 years ago
2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind
Alex777 [14]

The resultant speed of the plane  is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}

ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}

Hence, we have that the resultant speed of the plane  is (3) 226 m/s

Have a nice day!

5 0
3 years ago
A 6 cm object is 15 cm from a convex lens that has a focal length of 5 cm. What is the distance of the image from the lens, to t
Vesna [10]

Answer:

7.50 cm

Explanation:

The formula

1/v + 1/u = 1/f

Is used.

where.

u is the object distance.

v is the image distance.

f is the focal length of the lens.

1/v + 1/15 = 1/5

1/v = 1/5 - 1/15

1/v = (3-1)/15

1/v = 2/15

2v = 15

V = 15/2

V = 7.5 cm

For focal length, f in lens is always taken as negative for concave and positive for convex. ... And for image distance, V in lens it is taken as positive in Convex lens since image is formed on +X side. It is taken as negative in Concave lens since image is formed in -X side of the Cartesian.

4 0
3 years ago
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