If we increase the distance traveled when doing work and keep all other factors the same what will happen

In February 1955, a paratrooper fell 362 m from an airplane without being able to open his chute but happened to land in snow, s

serious [3.7K]

Answer:

s = 0.9689 m

Explanation:

given,

Height of fall of paratroopers = 362 m

speed of impact = 52 m/s

mass of paratrooper = 86 Kg

From from snow on him = 1.2 ✕ 10⁵ N

now using formula

F = m a

a = F/m

$a = \dfrac{1.2 \times 10^5}{86}$

$a =1395.35\ m/s^2$

Using equation of motion

v² = u² + 2 a s

$s =\dfrac{v^2}{2a}$

$s =\dfrac{52^2}{2\times 1395.35}$

s = 0.9689 m

The minimum depth of snow that would have stooped him is s = 0.9689 m

8 0

3 years ago

if a Firebird travels at a velocity of 0 to 60 mph in four seconds traveling east what was the acceleration of the Firebird

Tresset [83]

Answer:

6.7 m/s^2

Explanation:

The formula of acceleration is:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{v_2 - v_1}{t_2-t_1}}$

where $\displaystyle{\vec{a}}$ is acceleration, $\displaystyle{\vec{v}}$ is velocity and $\displaystyle{t}$ is time. $\displaystyle{v_2}$ means final velocity. $\displaystyle{v_1}$ means initial velocity, $\displaystyle{t_2}$ means final time and $\displaystyle{t_1}$ means initial time.

We are given that the Firebird travels at velocity of 0 to 60 mph in four seconds. Therefore:

Our initial velocity starts at 0 mph.
Our final velocity is at 60 mph.
Our initial time is 0 second.
Our final time is 4 seconds.

Since it travels to the east then our vector will be positive. However, acceleration has to be in m/s^2 unit (Sl unit) so we'll have to convert from mph (miles per hours) to m/s (meters per second) first.

We know that:

A mile equals to 1609.344 meters.
An hour equals to 60 minutes which a minute equals to 60 seconds. So 60 minutes will equal to 3600 seconds.

Now we divide 1609.344 by 3600 to find a unit rate of m/s:

$\displaystyle{\dfrac{1609.344}{3600} \ \, \sf{m/s}}\\\\\displaystyle{= 0.44704 \ \, \sf{m/s}}$

Now multiply 0.44704 m/s by 0 and 60 to get velocity in m/s unit:

Initial velocity = 0 m/s
Final velocity = 60 * 0.44704 = 26.82 m/s

Time is already in second so no need for conversion. Substitute known information in the formula:

$\displaystyle{\vec{a} = \dfrac{26.82-0}{4-0}}\\\\\displaystyle{\vec{a} = \dfrac{26.82}{4}}\\\\\displaystyle{\vec{a} = 6.7 \ \, \sf{m/s^2}}$

Therefore, the Firebird will accelerate at the rate of 6.7 m/s^2.

3 0

2 years ago

A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

expeople1 [14]

Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

given data

ship to travel = 1.7 × $10^6$ kilometers

turn = 70°

travel an additional = 2.7 × $10^6$ kilometers

solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

Qx =2.7 × $10^6$ cos(70)

Qy= 2.7 × $10^6$ sin(70)

so that

Hx = Px + Qx ............1

Hx = 2.62 × $10^6$

and

Hy = Py + Qy ..........2

Hy = 2.53 × $10^6$

so Magnitude = $\sqrt{((2.62 \times 10^6 )^2+(2.53 \times 10^6)^2)}$

Magnitude = 3.64 ×

so direction will be

tan စ = Hy ÷ Hx ......................3

tan စ = $\frac{2.53}{2.62}$

tan စ = 0.9656

စ = 43.9°

3 0

3 years ago

An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat

Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, $v_{max}$ = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2$

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s$

Therefore, the maximum speed is 21.39 m/s.

3 0

3 years ago

which of the following is not a projectile? a.a satellite b.a throw ball c.a ball on the ground d.a soaring arrow

Delvig [45]

C a ball on the ground

7 0

3 years ago

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