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masha68 [24]
2 years ago
14

An object is dropped from a 15 m ledge. How fast it is moving just before it hits the ground?

Physics
2 answers:
loris [4]2 years ago
7 0

Answer:

v_f = 17.15 m/s

Explanation:

Since the ball is dropped from the height

h = 15 m

so here initial speed of the ball is

u = 0

acceleration of the ball due to gravity is given as

a = 9.8 m/s^2

now the final speed of the ball is given as

v_f^2 - v_i^2 = 2 a d

v_f^2 - 0 = 2(9.8)(15)

v_f = 17.15 m/s

Sergeeva-Olga [200]2 years ago
4 0
By v^2 = u^2 + 2gh 
v^2 = 0 + 2 x 9.8 x 15
v = √294
v = 17.15 m/s 
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In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water d
Alexxandr [17]

Answer:

a) a = g / 3

b) x (3.0) = 14.7 m

c) m (3.0) = 29.4 g

Explanation:

Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

                             x*g = x * \frac{dv}{dt} + v^2

- Where,                v = Speed of the raindrop

- Proposed solution to given ODE:

                             v = a*t

Where,                  a = acceleration of raindrop

Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

- We know that acceleration (a) is the first derivative of velocity (v):

                             a = dv / dt   ... Eq 1

- Similarly, we know that velocity (v) is the first derivative of displacement (x):

                            v = dx / dt  , v = a*t ... proposed solution (Eq 2)

                             v .dt = dx = a*t . dt

- integrate both sides:

                             ∫a*t . dt = ∫dt

                             x = 0.5*a*t^2  ... Eq 3

- Substitute Eq1 , 2 , 3 into the given ODE:

                            0.5*a*t^2*g = 0.5*a^2 t^2 + a^2 t^2

                                                = 1.5 a^2 t^2

                            a = g / 3

- Using the acceleration of raindrop (a) and t = 3.00 second and plug into Eq 3:

                           x (t) = 0.5*a*t^2

                           x (t = 3.0) = 0.5*9.81*3^2 / 3

                           x (3.0) = 14.7 m  

- Using the relation of mass given, and k = 2.00 g/m, determine the mass of raindrop at time t = 3.0 s:

                           m (t) = k*x (t)

                           m (3.0) = 2.00*x(3.0)

                           m (3.0) = 2.00*14.7

                           m (3.0) = 29.4 g

6 0
2 years ago
If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?
Ghella [55]

<u>Given that</u>

mass (m) = 1300 Kg ,

height (h) = 1500 m

Determine the potential energy ?

     P.E = m × g × h

           = 1300 × 9.81 × 1500

           = 19129500  Joules

           = 19129.5 KJ

3 0
2 years ago
a falling skydiver slows from a speed of 52 m/s to 8 m/s in 0.8 sec as the parachute opens. what is the diver's acceleration and
Cloud [144]
I found the answers here. Hope this helps you! https://1.cdn.edl.io/sJTle6yxt3qVq7jHfdHRZJ3Xogj7ps6swBO9umNcZ6PO3SMN.docx
8 0
2 years ago
You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

7 0
2 years ago
A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is
Arte-miy333 [17]
Cuál es tu pregunta
7 0
1 year ago
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