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4 years ago

John traveled East at 10 m/s for ten

Physics

1 answer:

NNADVOKAT [17]4 years ago

5 0

Answer:

Option D

130 m

Explanation:

From the concept of speed, distance=speed*time

When traveling East, displacement=10*10=100 m (since he takes 10 seconds while traveling at a speed of 10 m/s)

When traveling West, displacement=5*6=30 m (since John takes 6 seconds to travel at a speed of 5 m/s)

Total displacement=Displacement East+ Displacement West=100+30=130 m

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a length of wire is cut into five equal pieces. the five pieces are then connected in parallel, with the resulting resistance be

Natali [406]

The equivalent resistance of n resistors connected in parallel is given by
$\frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n}$ (1)

In our problem, the resulting resistance of the 5 pieces connected in parallel is $R_{Eq}=2.00 \Omega$ , and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
$\frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R}$
From which we find $R= 5 \cdot 2 \Omega = 10 \Omega$ .

So, each piece of wire has a resistance of $10 \Omega$ . Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
$R_{Eq}=R_1 + R_2 + ...+R_n$
So if we apply it at our case, we have
$R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega$

therefore, the resistance of the original wire was $50 \Omega$ .

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3 years ago

Is the distance a baseball travels in the air after being hit distance a baseball travels in the air after being hit a discrete

Nookie1986 [14]

Answer:

Continuous random variable

Explanation:

The distance that baseball travels after being hit is a random variable and it assume any real value defined on the sample space.

The distance is measurable and thus is continuous random variable because continuous variable cannot be counted but could be measured.

3 0

3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$

W = 83 N

m = $\frac{83}{9.81}$

= 8.46

Now, we will balance the forces along the y-component as follows.

N = W + $F_{2}$

= 83 + 25 = 108 N

Now, balancing the forces along the x component as follows.

$F_{1} - F_{r}$ = ma

$20 - F_{r} = 8.46 \times (-0.9)$

$F_{r}$ = 7.614 N

Also, we know that relation between force and coefficient of friction is as follows.

$F_{r} = \mu \times N$

$\mu = \frac{F_{r}}{N}$

= $\frac{7.614}{108}$

= 0.0705

Thus, we can conclude that the coefficient of kinetic friction between the box and the floor is 0.0705.

7 0

4 years ago

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For an electron, magnitude of force on it is Select one • Bev bev Be BIL

maw [93]

Answer:

F = Bev

Explanation:

B is magnetic field density

e is the electron charge

v is the electron velocity

6 0

3 years ago

A 165 pound football player assumed a 4 point stance at the line. If his base of support was 30” wide, 38” deep, with the center

Hatshy [7]

Answer:

(1) The front attack is = 1528.56 inch-pounds (2) From the side is = 1237.5 inch- pounds. (3) from the rear is = 1528.32 inch pounds

Explanation:

Solution

Given

The weight of a football player is = 65 pound

Instance = 4 point

Now,

Let us consider that the weight on both hands are the same and weight on both knees are also same.

Let say, weight on each hand be declared as m1, and weight of each leg be m2

Then,

m₁+ m₁ + m₂ +m₂ = 165 pounds

m₁ + m₂ = 82.5 ---------( equation 1)

On the coordinate plane let us assume that the center of gravity G is at the origin (0,0)

so,

2m₁x₁ + 2m₂x₂ /2m₁ + 2m₂ = 0

m₁ * 16 - m₂ * 22 = 0

m₂ = 8 /π m₁------ (2)

Now, let substitute 8 /π m for m₂ in equation 1

m₁ + 8 /π m₁ = 82.5

where m₁ = 47.76 pounds and m₂ = 34.74 pounds

Now,

(a) The front attack

The weight in front that is the one arm is displaced and about the center of gravity

so,

the inch of pound needed is denoted as:

Mfront = 2m₂ * x₂ = 2 * 34.74 *22

The Mfront becomes = 1528.56 inch-pounds

(2) From the side

The weight on one leg and one hand is displaced about the center of gravity G

Mside = 2/2 * y (m₁ +m₂) = 15¹¹ * (82.5)

so,

Mside = 15 * 82.5

Mside = 1237.5 inch- pounds

Now,

For the real attack, the weight in rear end on the knees is displaced about the center of gravity

Mrear = 2m₁ * x₂ = 2 * 47.76 * 16

Therefore the Mrear = 1528.32 inch pounds

4 0

3 years ago