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My name is Ann [436]
3 years ago
6

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

splacement do you have after this walk?
Physics
1 answer:
xeze [42]3 years ago
5 0

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

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Explanation:

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2 years ago
A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer
Mrac [35]

Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.  

8 0
3 years ago
What has research determined about the orbit of an electron around a nucleus?
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4 0
3 years ago
A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) W
Anarel [89]

Answer:

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Explanation:

Given that

V_1=21 m/s,V_2=9.8m/s

We know that impulse(J) impulse is a vector quantity

J=P_2-P_1

We know that P=mV

SoJ=m(V_2-V_1)

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

F=\dfrac{dP}{dt}

or we can say that

F=m\left(\dfrac{v_2-V_1}{dt}\right)

So F=\dfrac{40.04}{0.0156}

F=2566.66 N

3 0
2 years ago
Why does a catalyst cause a reaction to proceed faster??
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The activation energy is lowered only
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There are more collisions per second only
The activation energy is lowered
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