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aleksklad [387]
2 years ago
13

A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in

Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873

(b) What is its acceleration once it starts to move, if that force is maintained?


m/s2

Physics
1 answer:
Novay_Z [31]2 years ago
4 0

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

<h3>Minimum force to be applied </h3>

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

  • μ(s) is coefficient of static friction
  • m is mass of the block
  • g is acceleration due to gravity

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

<h3>Acceleration of the block</h3>

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

Learn more about minimum force here: brainly.com/question/14353320

#SPJ1

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Early cameras were little more than a box with a pinhole on the side opposite the film. (a) What angular resolution would you ex
ollegr [7]

Answer:

angular resolution = 0.07270° = 1.269 × 10^{-3} rad

greatest distance from the camera = 118.20 m = 0.118 km

Explanation:

given data

diameter = 0.50 mm = 0.5 × 10^{-3} m

distance apart = 15 cm =  15× 10^{-2} m

wavelength λ = 520 nm = 520 × 10^{-9} m

to find out

angular resolution and greatest distance from the camera

solution

first we expression here angular resolution that is

sin θ = \frac{1.22* \lambda }{D}   .......................1

put here value λ is wavelength and d is diameter

we get

sin θ = \frac{1.22*520*10^{-9}}{0.5*10^{-3}}

θ = 0.07270° = 1.269 × 10^{-3} rad

and

distance from camera is calculate here as

θ = \frac{I}{r}    .................2

I = \frac{15*10^{-2}}{1.269*10^{-3}}

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3 years ago
.a car increases its speed from 10 m/s to 20/s in 2.5. seconds calculate it’s acceleration
Mekhanik [1.2K]

Answer:

\boxed {\boxed {\sf 4\ m/s^2}}

Explanation:

Acceleration is the rate of change of an object with respect to time. It is the change in velocity over the time.

a= \frac{v_f-v_i}{t}

The car starts at 10 meters per second and increases to 20 meters per second in 2.5 seconds.

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Substitute the values into the formula.

a= \frac{ 20 \ m/s - 10 \ m/s}{2.5 \ s }

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a= \frac{10 \ m/s}{2.5 \ s}

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