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Zinaida [17]
2 years ago
6

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

f 200 N. The box moves because of the two forces applied to it. Leon records the forces and direction of the forces acting on the box in his lab notebook. A 2 column table with 5 rows. Column 1 is labeled Force with entries Normal, Tension by Eduardo, Tension by Clara, Kinetic friction, Gravity. Column 2 is labeled Direction of Force Vector with entries Upward, Left, Right, Left, Downward. In the table, which force has the wrong direction? Tension by Eduardo Tension by Clara Kinetic friction Gravity
Physics
1 answer:
sergejj [24]2 years ago
8 0

Answer:

kinetic friction or something else

Explanation:

Draw a diagram to illustrate the problem, as shown below.

m = mass of the box, kg

mg =  weight of the box, N, acting downward

Because of Newton's 3rd law, a normal force of N = mg acts upward on the box.

C = 200 N, the force applied by Clara, horizontally to the right

E = 500 N, the force applied by Eduardo, horizontally to the left.

Because E is greater than C, the box will move left.

The frictional force, F, will resist the initial motion. Its magnitude is

F = μN = μmg, where μ = kinetic coefficient of friction.

F acts horizontally to the right.

From the given table, we can conclude that

Normal force acting upward is correct

Tension by Eduardoacting left is correct.

Tension by Clara acting right is correct.

Kinetic friction acting left is Incorrect.

Gravity acting downward is correct.correct

Answer:

The direction of the kinetic friction force is incorrect

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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0
2 years ago
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If vx=9.80 units and vy=-6.40 units, determine the magnitude and direction of v
dexar [7]
The resultant vector can be determined by the component vectors. The component vectors are vector lying along the x and y-axes. The equation for the resultant vector, v is:

v = √(vx² + vy²)
v = √[(9.80)² + (-6.40)²]
v = √137 or 11.7 units
5 0
2 years ago
The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu
IgorLugansk [536]

Answer:

g = 0.905 gE

W  = 67.9 N

Explanation:

given data

mass of Venus mv =  81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = \frac{Gm}{R^2}   = 9.80 m/s²  ............1

so

g = \frac{G(0.815)}{0.949R^2}  

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W  = 0.905 × 75

W  = 67.9 N

7 0
2 years ago
You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current i
Anni [7]

Answer:

<em>radius of the loop =  7.9 mm</em>

<em>number of turns N ≅ 399 turns</em>

Explanation:

length of wire L= 2 m

field strength B = 3 mT = 0.003 T

current I = 12 A

recall that field strength B = μnI

where n is the turn per unit length

vacuum permeability μ  = 4\pi *10^{-7}  T-m/A = 1.256 x 10^-6 T-m/A

imputing values, we have

0.003 = 1.256 x 10^−6 x n x 12

0.003 = 1.507 x 10^-5 x n

n = 199.07 turns per unit length

for a length of 2 m,

number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>

since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

this length is also equal to the circumference of each loop

the circumference of each loop = 2\pi r

0.005 = 2 x 3.142 x r

r = 0.005/6.284 = 7.9*10^{-4} m = 0.0079 m =<em> 7.9 mm</em>

8 0
2 years ago
A sample of silver (with work function Φ=4.52 eV ) is exposed to an ultraviolet light source (????=200 nm), which results in the
Crank

Answer:

1. c

2. e

3. d

Explanation:

1.

From Einstein's Photoelectric Equation, we know that:

Energy given up by photon = Work Function + K.E of Electron

hc/λ = φ + K.E

where,

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m

φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J

Therefore,

(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E

K.E = (9.939 - 8.16) x 10⁻¹⁹ J

K.E = 1.778 x 10⁻¹⁹ J

The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:

c. <u>less energetic photo-electrons (on average)</u>

<u></u>

2.

The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,

e. <u>more photo-electrons ejected</u>

<u></u>

3.

X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.

d. <u>more energetic photo-electrons (on average)</u>

8 0
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