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Cloud [144]
3 years ago
10

a man is pulling a block with a mass of 6.2kg across a horizontal surface and accelerates the block at a rate of 0.50m/s.the coe

fficient of kinetic friction between the block and the surface is 0.24. what is the magnitude with which the man pulls?
Physics
2 answers:
Ivahew [28]3 years ago
8 0

Answer:

17.68 N

Explanation:

mass of block, m = 6.2 kg

acceleration, a = 0.5 m/s^2

coefficient of friction, μ = 0.24

Net force - mass x acceleration

F - f = ma

where, F is the pulling force, f is the friction force and a be the acceleration

F - μ mg = ma

F - 0.24 x 6.2 x 9.8 = 6.2 x 0.5

F - 14.58 = 3.1

F = 17.68 N

Thus, the pulling force is 17.68 N.

klasskru [66]3 years ago
7 0
Mg = 6.2 x 9.81 = 60.822 This is also normal force. Coefficient of friction times normal force is the force due to friction: 60.822 x 0.24 = 14.6N F = MA so F(your force) - F(friction) = 6.2 x 0.5 = 3.1 Your answer is 3.1+ 14.6 I hope this is correct though I might be wrong.
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5 0
3 years ago
A toy car has an initial acceleration of 2 m/s" across a horizontal surface after it is released from rest. After the car travel
Ganezh [65]

Answer:

Closed system, because the speed of the car is as expected in the case where an object has uniform acceleration for a time t

Explanation:

Here in the question it is mentioned that a toy car has an initial acceleration of 2m/s²  across a horizontal surface so we can say that it is acted upon by an external force

Assuming that the acceleration is constant and the reason for this assumption is there at the last

The major difference between an open system and closed system is in case of open system there will be transfer of matter and in case of closed system there will be no change in matter of the system

If acceleration is constant in case of closed system we can expect the speed of the car after a time t by using the formula

  s = u×t + 0·5×a×t²

where s is the distance travelled

t is the time taken to travel that distance

u is the initial velocity

a is the acceleration of that system

But in case of open system as there will be a change of mass there will be a change in velocity of the system so in this case we cannot expect the speed of the car after a time t

And if the acceleration is not constant then we cannot say that the toy car is an open system or closed system, that is why we are assuming that the acceleration of the toy car is constant

3 0
3 years ago
Read 2 more answers
Usually, when the temperature is increased, what will happen to the rate of dissolving?
ArbitrLikvidat [17]
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5 0
3 years ago
A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0
3 years ago
The length of a simple pendulum is 0.66 m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree
lina2011 [118]
A) the periodic time is given by the equation;
 T= 2π√(L/g)
For the frequency will be obtained by 1/T (Hz)
T = 2 × 3.14 √ (0.66/9.81)
   = 6.28 × √0.0673
    = 1.6289 Seconds
Frequency = 1/T = f = 1/1.6289
 thus; frequency = 0.614 Hz

b)  The vertical distance, the height is given by
 h= 0.66 cos 12
 h = 0.65 m
Vertical fall at the lowest point = 0.66 - 0.65 = 0.01 m
Applying conservation of energy
energy lost (MgΔh) = KE gained (1/2mv²)
 mgh = 1/2mv²
  v² = 2gΔh = 2×9.81 × 0.01 
                   = 0.1962
v = 0.443 m/s

c) total energy = KE + GPE = KE when GPE is equal to zero (at the lowest point possible)
Thus total energy is equal to;
E = 1/2mv²
   = 1/2 × 0.310 × 0.443²
   = 0.0304 J


4 0
3 years ago
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