How long will it take to go 150,000 m traveling at 50 km/hr

If a freely falling rock were equipped with a speedometer, by how much would itd speed readings increase with each second if it

Len [333]

" 20 m/s² " means that if gravity is the ONLY force on the object
(the object is in 'free fall'), then its speed increases by 20 m/s
every second.

That's the answer to your question. Now, let me ask you
another one:

How does a speedometer tied to a falling rock work ?
How can it measure the rock's speed ?

Maybe one way would be to have a little tiny propeller on
the front of the speedometer, and it could measure how fast
the propeller is spinning as the rock falls through the air ?
Great idea. But we already said the rock is in free-fall,
so there's no air resistance, we can't have any air, and
there's nothing to spin the propeller.

How would you do it ? How can you measure the rock's speed ?

3 0

4 years ago

Which element is this?

hammer [34]

Answer:Silicon

Explanation:

You add up the protons and neutrons and then look at the element paper

7 0

3 years ago

Read 2 more answers

The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approa

Oduvanchick [21]

Answer:

The boat is approaching the dock at a rate of <u>2.5 ft/s</u>.

Explanation:

Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.

Given:

The height of dock above water (h) = 6 feet

Rate of pull of rope or rate of change of rope is, $\frac{dl}{dt}=2\ ft/s$

As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.

Now, the above situation represents a right angled triangle as shown below.

Using Pythagoras Theorem, we have:

$l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)$

Now, differentiating the above equation with time 't', we get:

$2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)$

Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,

$b^2=10^2-36\\\\b=\sqrt{64}=8\ ft$

Now, substituting all the given values in equation (2) and solve for $\frac{db}{dt}$ . This gives,

$\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s$

Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.

7 0

3 years ago

A block of mass m=16.8 kg is sliding on a surface with initial velocity v=23.2 m/s. The block has a coefficient of kinetic frict

Firdavs [7]

Answer:

t = 23.92 s

Explanation:

Newton's second law:

∑F = m*a Formula (1)

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

m : mass s (kg)

a : acceleration (m/s²)

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

Forces acting on the block

W: Weight of the block : In vertical direction, downward

FN : Normal force : perpendicular to the floor, upward

fk : Kinetic friction force: parallel to the floor and opposite to the movement

F = 86.4 N , in the direction of the motion

Calculated of the W

W= m*g

W= 16.8 kg* 9.8 m/s² = 164.64 N

Calculated of the FN

We apply the formula (1)

∑Fy = m*ay ay = 0

FN - W = 0

FN = W

FN = 164.64 N

Calculated of the fk

fk = μk*FN

fk = 0.426* 164.64 N

fk = 70.13 N

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

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

F - fk = m*a

86.4 -70.13 = (16.8)*(-a)

16.26 = (16.8)*(-a)

a = -(16.26 )/ (16.8)

a = - 0.97 m/s²

Kinematics of the block

Because the block moves with uniformly accelerated movement we apply the following formula :

vf = v₀ + a*t Formula (2)

Where:

t: time interval (m)

v₀: initial speed (m/s)

vf: final speed (m/s)

Data:

v₀ = 23.2 m/s

vf = 0

a = -0.97 m/s²

Time it takes for the block to stop

We replace data in the formula (2) to calculate the time

vf= v₀+a*t

0 = 23.2+( -0.97)*t

(0.97)*t = 23.2

t = 23.2 / (0.97)

t = 23.92 s

7 0

3 years ago

HELP!!! I will 5 star if its helpful :(

victus00 [196]

Answer:

Net Force Formula: F1+F2+F3...+FN

*(MAKE SURE YOU MAKE NOTE OF THE NEGATIVE FORCES AND SUBTRACT THEM!)*

Explanation:

Top left: 17N+25N+25N-42N= 25N

Top right: 65N+200N-65N-150N-200N= 150N

Bottom left: 189N+123N+284N-96N-188N-312N= 0N

Bottom right: 34N+77N-12N-34N= 65N

I hope this helps! :))

5 0

3 years ago