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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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The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

6 0
3 years ago
You want to slide a 0.39 kg book across a table. If the coefficient of kinetic friction is .21, what force is required to move t
uranmaximum [27]
Find the force that would be required in the absence of friction first, then calculate the force of friction and add them together.  This is done because the friction force is going to have to be compensated for.  We will need that much more force than we otherwise would to achieve the desired acceleration:

F_{NoFric}=ma=0.39kg \times0.18 \frac{m}{s^2}  =0.0702N

The friction force will be given by the normal force times the coefficient of friction.  Here the normal force is just its weight, mg

F_{Fric}=0.39kg \times 9.8 \frac{m}{s^2} \times 0.21=0.803N

Now the total force required is:

0.0702N+0.803N=0.873N

5 0
3 years ago
Help ASAP PLEASE.............:3​
Evgesh-ka [11]

Answer:

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8 0
3 years ago
What is the equation for frequency, wavelength, and speed of a wave?
Nastasia [14]

Answer:

       λ = v/f

Explanation:

frequency=f

wavelength = λ

speed of a wave=v

7 0
3 years ago
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snow_tiger [21]

Answer:

M.A = load/ effort

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= 3

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40cm/10cm

=4

efficiency= 3/4*100

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4 0
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