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Lena [83]
3 years ago
8

A net force of 1.0N acts on a 4.0 kg object, initially at rest, for 4.0 seconds. What is the distance the object moves during th

e same time?
Physics
1 answer:
Eva8 [605]3 years ago
8 0
Use F=ma to solve for a=0.25m/s^2. Now to get distance you need to integrate this twice to get
x=(1/2)at^2. Plug in a and t to get
x=2m
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An ant is crawling up a leaf that is hangin vertically. it lifts 0:007 kg crumb up 0:12 meters. to do this, the ant exerts 0.069
Afina-wow [57]

Answer:

.000828 j

Explanation:

Work = F * d

          .0069 N * .12 m = .000828 j

5 0
2 years ago
Why are hurricanes considered more damaging than tornadoes when tornadoes have stronger winds
chubhunter [2.5K]

-- A tornado follows a path that's a few miles wide, for a few hours.
   Then it's all over.

-- A hurricane follows a path that's several hundred miles wide,
   for a week or two, before it's over.
   Then comes the rain, continuing on the same path, for another week.

8 0
3 years ago
Read 2 more answers
A box is pulled along a floor by a force of 3.0 N. The friction acting on the box is 1.0 N, as shown. How much kinetic energy do
Furkat [3]

Answer:

4 J

Explanation:

From the image attached, we can see 2 horizontal forces acting on the box albeit in opposite directions.

Now, the net force will be;

F_net = 3 - 1

F_net = 2 N

To move a distance of 2 metres, kinetic energy is;

K.E = Force × Distance = 2 × 2 = 4 J

7 0
3 years ago
A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.
fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

(c) 352.9 Hz

Explanation:

For an open pipe,  the velocity is given by

v=\frac {2Lf}{n}

Making L the subject then

L=\frac {nV}{2f}

Where f is the frequency,  L is the length,  n is harmonic number,  v is velocity

Substituting 1 for n,  320 Hz for f and 331 m/s for v then

L=\frac {1*331}{2*320}=0.5171875\approx 0.52m

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, f2=2\times \frac {v}{2L} and f3=3\times \frac {v}{2L}

f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz

(c)

When v=367 m/s then

f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz

5 0
3 years ago
Two equal magnitude electric charges are separated by a distance d. The electric potential at the midpoint between these two cha
Ann [662]

Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

8 0
3 years ago
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