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matrenka [14]
2 years ago
15

If a boulder has a weight of 675,000 N what is it’s mass

Physics
1 answer:
Anarel [89]2 years ago
6 0

Weight = (mass) x (acceleration of gravity where the object is)

You didn't tell us WHERE the boulder is, so I have to assume that it's on Mars, where the acceleration of gravity is 3.71 m/s².

675,000 N = (mass) (3.71 m/s²)

Mass = (675,000 N) / (3.71 m/s²)

<em>Mass = 181,941 kilograms</em>

The same weight on Earth would suggest a mass of only 68,807 kg, so you can see how important it is to know where you are when you make your measurements.

You might be interested in
Which waves have oscillations parallel to their direction of motion
wolverine [178]

Answer:

The direction of these oscillations is the difference between longitudinal or transverse waves. In longitudinal waves, the vibrations are parallel to the direction of wave travel. In transverse waves, the vibrations are at right angles to the direction of wave travel.

Explanation:

7 0
2 years ago
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
White raven [17]

Answer:

400ft.    32ft/s       -32ft/s

Explanation:

In reality the gravitational acceleration is 9.81 so the quadratic coefficient of the function should be 9.81/2

Anyway for the sake of assumtion let us takes=160t-16t^2

 

ds/dt=160-32t=0

 

t=160/32= 5 seconds.

s=160*160/32-16*(160/32)^2= 400 mts

 

 

s=384 mts

160t-16t^2=384

i.e

16t^2-160t+384=0

 

t^2-10t+24=0

(t-6)(t-4)=0

t=[4,6]

we have to take t=4 because it is all the up i.e <5

 

velocity =v=ds/dt=160-32t

 

v=160-32*4=32 ft/sec still going up

 

for all the way down take t=6 whuch is >5

 

v=160-6*32=-32 ft/sec (falling down!!!)

6 0
2 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
How do very small objects behave?
nataly862011 [7]
A. very small objects behave like like particles.
7 0
3 years ago
Read 2 more answers
Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express
AlladinOne [14]

Answer:

Vb = k Q / r        r <R

Vb = k q / R³ (R² - r²)    r >R

Explanation:

The electic potential is defined by

             ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

             VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

           Va = - ∫ (k Q / r²) dr

           -Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

          Vb = k Q / r        r <R

         

We perform the calculation of the power with the expression of the electric field that they give us

           Vb = - int (kQ / R3 r) dr

  We integrate and evaluate from the starting point r = R to the final point r <R

         Vb = ∫kq / R³ r dr

         Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0
3 years ago
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