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2 years ago

Describe the forces that act on a skydiver before

Physics

1 answer:

MrRissso [65]2 years ago

8 0

Answer:

<em>Before the parachute opens: Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.</em>

<em>Once the parachute is opened, the air resistance overwhelms the downward force of gravity. The net force and the acceleration on the falling skydiver is upward. An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.</em>

<h3>I HOPE THIS WILL HELP YOU IF NOT THEN SORRY</h3>

HAVE A GREAT DAY :)

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Match each statement with the characteristic of scientists.

Dimas [21]

Answer:

Options B, A, D, C

Explanation:

When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.

5 0

3 years ago

Read 2 more answers

A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld

lesantik [10]

Answer: The magnitude of the proton's acceleration is 0.748 ×10^14 m/s²

Explanation:
the velocity ,v, of ththe proton = 5.02×10^6 m/s
Magnitude , B , of the magnetic field = 0.180 T

First , we need to find the magnitude of the Force on the proton. This is given by the relation :
F = q(v x B) = qvBsinθ

where 'q' is the charge of proton , q= 1.6×10^-19 C
θ is the angle the proton makes with the direction of the magnetic field

Putting the respective values of v, B ,θ in the above equation, we get:

F = (1.6×10^-19 C)(5.02×10^6 m/s)(0.180T) sin60°
∴ F = 1.25 ×10^-13 N

Now , from Newton's second law we know that ,
F=m×a

∴ a = F/m

Mass of a proton = 1.67×10^27 kg
a= 1.25 × 10^-13 N / 1.67 × 10^27 kg

a= 0.748 × 10^14 m/s² =acceleration of the proton

(To know more about Magnetic Fields : brainly.com/question/9095546)

5 0

1 year ago

A raindrop of radius r falls from a certain height h above the ground. The work done by

lisabon 2012 [21]

Answer:

fhc

Explanation:

chchfufuufufyfydyedhxhfud wyruficjc

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2 years ago

A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa

Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

$Q=P\tan\phi$ ...(I)

We need to calculate the $\phi$

Using formula of $\phi$

$\phi=\cos^{-1}(Power\ factor)$

Put the value into the formula

$\phi=\cos^{-1}(0.6)$

$\phi=53.13^{\circ}$

Put the value of Φ in equation (I)

$Q=600\tan(53.13)$

$Q=799.99\ kVAR$

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

$Q'=600\tan(0.95)$

$Q'=9.94\ KVAR$

We need to calculate the difference between Q and Q'

$Q''=Q-Q'$

Put the value into the formula

$Q''=799.99-9.94$

$Q''=790.05\ KVAR$

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0

3 years ago

Strontium−90 is one of the products of the fission of uranium−235. This strontium isotope is radioactive, with a half-life of 28

Ludmilka [50]

Answer : The time passed in years is 20.7 years.

Explanation :

Half-life = 28.1 years

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{28.1\text{ years}}$

$k=2.47\times 10^{-2}\text{ years}^{-1}$

Now we have to calculate the time passed.

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $2.47\times 10^{-2}\text{ years}^{-1}$

t = time passed by the sample = ?

a = initial amount of the reactant = 1.00 g

a - x = amount left after decay process = 0.600 g

Now put all the given values in above equation, we get

$t=\frac{2.303}{2.47\times 10^{-2}}\log\frac{1.00}{0.600}$

$t=20.7\text{ years}$

Therefore, the time passed in years is 20.7 years.

8 0

3 years ago