All categories

3 years ago

If a body travels half its total path in the last 1.50 s of its fall from rest, find the total time of its fall (in seconds).

Physics

1 answer:

svetoff [14.1K]3 years ago

6 0

Answer:

time to fall is 3.914 seconds

Explanation:

given data

half distance time = 1.50 s

to find out

find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at² .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1

and

velocity will be

v = u + at

so velocity v = 0+ 9.8(t-1.5) ..................2

so first we find time

0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)

solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

You might be interested in

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

Solving:

STatiana [176]

Answer:

1. 31,536,000 seconds

2. Car B traveled a longer distance

3. Volume of box = 0.235887 cubic meters

Explanation:

Q1. Age in seconds

1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Therefore 1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

In scientific notation this would be $3.1536\times 10^7 \text{ seconds}$

Q2. Comparing km and miles

Given:
1 km = 1000m and 1 m = 3.3ft, I km = 1000 x 3.3 = 3300 ft.

Convert Car A distance of 25.7km to feet :
25.7 km. = 25. 7 x 3300 ft. = 84,810 ft.

For Car B that traveled 20 miles,
20 miles = 20 x 5280 = 105,600 ft.

Since 105,600 > 84,810, car B traveled a longer distance

Q3. Volume of wooden box

The wooden box is in the shape of a rectangular prism
It volume is L x W x H
Volume = 1.525 x 0.30 x 0.5156 = 0.235887 cubic meters

5 0

9 months ago

In your own words..Describe Friction Force..<br><br> In your own words..describe Applied force...

BARSIC [14]

I think force is something to do with a force being applied to a movement or is just to pull but not sure sorry if this aint right

5 0

3 years ago

Read 2 more answers

Which is true about the velocity of sound waves in solids compared to air

tankabanditka [31]

C. Travels slower in solids because the particles are closer together.
<span />

4 0

3 years ago

Read 2 more answers

2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting po

Sonja [21]

Answer:

a) t = 11.2 s

b) v = 70.5 mph

Explanation:

Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:

$t = \frac{v_{f} - v_{o}}{a} (1)$

where vf = 50 mph, and v₀ = 10 mph.
However, we still lack the value of a.
Assuming that the acceleration is constant, we can use the following kinematic equation:

$v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)$

Since we know that Δx = 500 ft, we could solve (2) for a.
In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:

$v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)$

$v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)$

We can do the same process with Δx, from ft to m, as follows:

$\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)$

Replacing (3), (4), and (5) in (2) and solving for a, we get:

$a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)$

Replacing (6) in (1) we finally get the value of the time t:

$t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)$

Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:

$v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)$

If we convert vf again to mph, we have:

$v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph (9)$

4 0

2 years ago