Ask question

Ask question

All categories

antiseptic1488 [7]

3 years ago

10

if the DIY parachute is dropped freely from the rooftop of the building of a building and reached the ground 3s later.Upon reach

ing the ground,What is the (a)final velocity of DYI parachutes ?(b) Height of the rooftop?

1 answer:

Goshia [24]3 years ago

5 0

Answer:

It depends on what the parachute is made out of.

Explanation:

You might be interested in

In order to relieve excessive pump pressure in an engine's internal oil system, most engines are equipped with a

svlad2 [7]

Answer:

A-Oil pressure relief valve.

5 0

2 years ago

A dentist’s drill starts from rest. After 1.46 sof constant angular acceleration, it turns at arate of 27000 rev/min.Find the dr

Black_prince [1.1K]

Answer:

616.3 rad/s²

Explanation:

Given that

t= 1.46 s

Initial angular velocity ,ωi = 0 rad/s

Final angular velocity ωf= 27000 rev/min

Angular speed in the rad/s given as

$\omega_f=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values

$\omega_f=\dfrac{2\times 27000}{60}\ rad/s$

ωf=900 rad/s

We know that (if acceleration is constant)

ωf=ωi + α t

α=Angular acceleration

900 = 0 + α x 1.46

$\alpha=\dfrac{900}{1.46}\ rad/s^2\\\alpha=616.43\ rad/s^2$

Therefore the acceleration will be 616.3 rad/s²

4 0

3 years ago

HELP MEEEEE HURRRYYYY EASYY

alisha [4.7K]

Answer:

C

Explanation:

thats what i think it depends on the amount of force someone dropped the ball with.

7 0

3 years ago

An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

$C = \frac{Q}{V} = e_0\frac{A}{d}$

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

$V = \frac{Q}{e_0} *\frac{d}{A}$

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

$V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V$

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0

3 years ago

Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts

podryga [215]

Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final height =?

$y_{0}$ : is the initial height = 1 m

$v_{0_{y}$ : is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time

First, we need to find the time by using the following equation:

$t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s$

Now, the height is:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m$

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

$0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}$

$1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0$

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

$x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m$

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!

4 0

3 years ago