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2 years ago

5

One of the great dangers to mountain climbers is an avalanche, in which a large mass of snow and ice breaks loose and goes on an

essentially frictionless "ride" down a mountainside on a cushion of compressed air. The acceleration of gravity is 9.8 m/s 2 . avalanche µ = 0 347 m 27.5 ◦ If you were on a 27.5 ◦ slope and an avalanche started 347 m up the slope, how much time would you have to get out of the way?

1 answer:

sammy [17]2 years ago

5 0

Answer:

Explanation:

The acceleration of an object down a slope (neglecting friction, µ = 0) is:

a = g × sin θ

Where,

g is the acceleration due to gravity and θ is the angle of the slope.

a = (9.8 × sin (21.5º)

= 3.592 m/s²

Using equations of motion,

S = ut + 1/2at²

Since, u = 0,

S = 1/2at²

347 = 1/2 × (3.592)t²

t² = 193.21

= sqrt(193.21)

= 13.9 s.

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

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2 years ago

Devonte pushes a wheelbarrow with 830 W of power. How much work is required to get the wheelbarrow across the yard in 11 s? Roun

xxMikexx [17]

Answer: 9130 joules

Explanation:

Workdone by wheelbarrow = ?

Time = 11 seconds

Power = 830 watts

Recall that power is the rate of doing work. Thus, power is workdone divided by time taken.

i.e Power = (workdone/time)

830 watts = Workdone / 11 seconds

Workdone = 830 watts x 11 seconds

Workdone = 9130 joules

Thus, 9130 joules of work is required to get the wheelbarrow across the yard.

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2 years ago

Read 2 more answers

A bike first accelerates from 0.0 m/s to 5.0 m/s in 4.5s, then continues at this constant speed for another 4.5 s. What is the t

Veseljchak [2.6K]

Assuming constant acceleration, the distance travelled in the first 4.5s is:

0.5*5.0*4.5 = 11.25m

The distance travelled in the next 4.5s is:

5.0*4.5 = 22.5m

The total distance travelled is:

11.25 + 22.5 = 33.75m

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2 years ago

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a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

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2 years ago

A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that μs = 0.50 and μk = 0

frutty [35]

Answer:

x = 1.63 m

Explanation:

mass (m) = 10 kg

μk = 0.3

velocity (v) = 3.1 m/s

Assuming that most of the computers weight is applied on the belt instantaneously, we can apply the constant acceleration equation below

x = $v^{2}/2a$

where a = μk.g , therefore

x = $v^{2}$ /2μk.g

x = (3.1 x 3.1)/(2 x 0.3 x 9.8)

x = 1.63 m

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2 years ago