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Sav [38]
4 years ago
11

If the man on the left pulls on the object with a force of 500 N and the man on the right pulls on the object with a force of 75

0 N, what is the net force exerted on the object?
A.
1250 N to the right
B.
1250 N to the left
C.
250 N to the right
D.
250 N to the left
Physics
1 answer:
Igoryamba4 years ago
3 0
Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.
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A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes.
Semenov [28]

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

\epsilon=NA\dfrac{dB}{dt}

For N = 1

\epsilon=A\dfrac{dB}{dt}

We need to calculate the current

Using formula of current

i=\dfrac{\epsilon}{R}

Put the value of emf

i=\dfrac{A\dfrac{dB}{dt}}{R}

Now, if the number of turn is 22 , then induced emf would be

\epsilon'=NA\dfrac{dB}{dt}

Then the current would be

i'=\dfrac{\epsilon'}{NR}

i'=\dfrac{NA\dfrac{dB}{dt}}{NR}

i'=\dfrac{A\dfrac{dB}{dt}}{R}

i'=i

Hence, The current would be same in both situation.

4 0
3 years ago
Put the pairs of atoms in order, with the pair that has the biggest electronegativity difference between the two atoms at the to
mojhsa [17]
Electronegativity is the measure of the tendency of an atom to attract a bonding pair of electrons. In the periodic table, electronegativity increase across the period because the charges on the nucleus increase. The correct arrangement for the atoms given above is as follows
Flourine and Francium
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5 0
3 years ago
You and some friends decide to take a canoe trip down the Wood River. The stretch of river you are traveling flows at a roughly
Anna35 [415]

Answer:

t= 1.2 hours

Explanation:

Define first di distance between the points, so

\bar{x}_{canoe}=2*(2+3)=10

\bar{x}_{water}=2*2=4

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d= \bar{x}_{canoe}- \bar{x}_{water}

d= 10-4 = 6miles

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5 0
3 years ago
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Softa [21]

Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

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Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

47.2(v_g,i) = 177(v_p,i)

Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

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Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

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5 0
3 years ago
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natulia [17]

Answer:

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Explanation:

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Let's set a reference system where the x axis is parallel to the road

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       N_B + N_A = W_van + W_load

X axis

     fr = ma

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the total mass is

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the friction force has the expression

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we substitute

      a = μ (W_van + W_load)    \frac{g}{W_van + W_load}

      a = μ g

taking the acceleration let's use the kinematic relations where the final velocity is zero

       v² = v₀² - 2 a x

       0 = v₀² -2a x

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        x = \frac{v_o^2}{2 \mu g}

        x = \frac{40^2}{2 \ 32 \  \mu}

        x = 25 / μ     [ ft]

5 0
3 years ago
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