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UkoKoshka [18]
1 year ago
11

A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa

lling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target?
Physics
1 answer:
elena55 [62]1 year ago
6 0

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

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A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

                                r₁ = 0.01 m

diameter of second piston,  d₂ = 0.15 m

                                r₂ = 0.075 m

mass of the jack to lift = ?

now,

    \dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}

    \dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}

    \dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}

    F_2= \dfrac{250}{0.01^2}\times {0.075^2}

               F₂ = 14062.5 N

F = m g

m = \dfrac{F}{g}

m = \dfrac{14062.5}{9.8}

m = 1435 Kg

hence, the correct answer is option B

5 0
2 years ago
A distance of 2.00 mm separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of
aleksklad [387]

Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:

F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2

And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)

The mass of the two objects are 24.96 kg

6 0
1 year ago
A man carries a load of mass 2.6kg from one end of a uniform pole 100cm long which has a mass 0.4kg. The pole rest on his should
miskamm [114]

Answer:

F = 39.2 N   (hand force) and    N = 68.6 N (shoulder force)

Explanation:

In this exercise we must use the rotational and translational equilibrium conditions, we have several forces: the weight (W) of the pole applied at its geometric center, the load (w1) at one end, the shoulder support (N) 60 cm from the load and hand force (F) at the other end of the pole

Let's set the reference system at the fit point of the shoulder

     ∑ τ = 0

We will assume that the counterclockwise turns are positive

    w₁ 0.60 + W 0.1 + F₁ 0 - F 0.4 = 0

all distances are measured from the support of the man (x₀ = 0.60 m)

    F = (w₁ 0.60 + W 0.1) / 0.4

    F = (m₁ 0.6 + m 0.1) g / 0.4

let's calculate

    F = (2.6 0.6 + 0.4 0.1) 9.8 / 0.4

    F = 39.2 N

this is the force that the hand must exert to keep the system in balance

We apply the translational equilibrium condition

    -w₁ -W + N - F = 0

     N = w₁ + W + F

     N = (m₁ + m) g + F

let's calculate

     N = (2.6 + 0.4) 9.8 + 39.2

     N = 68.6 N

6 0
1 year ago
A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?
Aleksandr [31]

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

8 0
2 years ago
Read 2 more answers
Among the largest passenger ships currently in use, the Norway has been in service the longest. The Norway is more than 300 m lo
LenaWriter [7]

Answer:

6.33\times 10^8\ kg\cdot m/s

Explanation:

Mass of the ship (m) = 6.9 × 10⁷ kg

Speed of the ship (v) = 33 km/h

First, let us convert the speed from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

So, 33 km/h = 33\times \frac{5}{18}=9.17\ m/s

Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.

So, here, length of the ship doesn't play any role in the determination of the momentum.

Magnitude of momentum of the ship = Mass × Speed

                                                             = (6.9\times 10^7\ kg)(9.17\ m/s)

                                                             = 6.33\times 10^8\ kg\cdot m/s

Therefore, the magnitude of ship's momentum is 6.33\times 10^8\ kg\cdot m/s.

6 0
2 years ago
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