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4 years ago

Which claim would most likely be considered valid?

Physics

2 answers:

Mekhanik [1.2K]4 years ago

7 0

Please state the options and I will answer to the best of my abilities XD

Alex4 years ago

5 0

"A wildlife biologist stating humans are destroying the migration" as it is the only one that could be proved with facts and not opinion

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A 72 kg skydiver can be modeled as a rectangular "box" with dimensions 21 cm × 41 cm × 170 cm . if he falls feet first, his drag

LuckyWell [14K]

Formula for terminal velocity is:

Vt = √(2mg/ρACd)
Vt = terminal velocity = ?
m = mass of the falling object = 72 kg
g = gravitational acceleration = 9.81 m/s^2
Cd = drag coefficient = 0.80
ρ = density of the fluid/gas = 1.2 kg/m^3
A = projected area of the object (feet first) = 0.21 m * 0.41 m = 0.0861 m^2

Therefore:

Vt = √(2 * 72 * 9.81 / 1.2 * 0.0861 * 0.80)

Vt = 130.73 m/s

4 0

3 years ago

А A van accelerates from Amst to zomst in 8s. How far does it travel in this time?

8_murik_8 [283]

Answer:

1663m

Explanation:

mark brainliest

6 0

3 years ago

A basketball player wishes to make a stunning last-second basket. With .8 seconds on the clock, she wants the ball to enter the

zheka24 [161]

Answer:

(1) $46.86^\circ$

(2) Diagram has been attached in the solution.

Explanation:

This question is from projectile motion.

From the given question, we will discuss the motion of the basket ball only in the vertical direction from which we will be able to find out the angle of the initial velocity with the horizontal with which it should be shoot to enter the hoop.

Part (1):

Let us assume:

$y_i$ = initial position of the basket ball = 2.1 m
$y_f$ = final position of the basket ball = 3.05 m
$a_y$ = acceleration of the ball along the vertical = $-9.8\ m/s^2$
$t$ = time taken to reach the goal = 0.8 s
$\theta$ = angle of the initial velocity with the horizontal
$u$ = initial speed of the ball = 7 m/s
$u_y$ = initial vertical velocity of the ball = u\sin \theta

Using the equation of motion for constant acceleration, we have

$y_f-y_i=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 3.05-2.1=u\sin \theta (0.8) +\dfrac{1}{2}\times (-9.8)(0.8)^2\\\Rightarrow 0.95=7\times \sin \theta (0.8) -3.136\\\Rightarrow 0.95=5.6\sin \theta -3.136\\\Rightarrow 5.6\sin \theta= 0.95+3.136\\\Rightarrow 5.6\sin \theta= 4.086\\\Rightarrow \sin \theta= \dfrac{4.086}{5.6}\\\Rightarrow \sin \theta=0.729\\\Rightarrow \theta=\sin^{-1}(0.729)\\\Rightarrow \theta=46.86^\circ$

Hence, the angle of the shoot of the basket ball with the horizontal is $46.86^\circ$ such that it reaches the hoop on time.

Part (2):

For this part, a diagram has been attached.

4 0

3 years ago

The electric field strength is 20,000 N/C inside a parallel-platecapacitor with a 1.0 mm spacing. An electron is released fromre

creativ13 [48]

Explanation:

It is given that,

Electric field strength, E = 20,000 N/C

Spacing between parallel-plate capacitor, d = 1 mm = 0.001 m

Initial velocity of electron, u = 0

Let v is the electron’s speed when it reaches the positive plate. The force acting on the electron is :

$F=qE$

Also, $ma=qE$

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 20,000}{9.1\times 10^{-31}}$

$a=3.51\times 10^{15}\ m/s^2$

Using third equation of motion as :

$v^2-u^2=2ad$

$v=\sqrt{2ad}$

$v=\sqrt{2\times 3.51\times 10^{15}\times 0.001}$

v = 2649528.2599 m/s

$v=2.64\times 10^6\ m/s$

So, the velocity of the electron when it reaches the positive plate is $2.64\times 10^6\ m/s$ . Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

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kozerog [31]

Newton's second equation of motion :-

S=ut+1/2at^2 [where, u is the initial velocity, a is the acceleration and t is the time interval]

This Equation simply finds a relation between distance travelled by a particle (classically) under uniform acceleration.

So let's see what pieces of information (bundles of equations) do we have with us, initially.

We have, a very primary equation with us,

dS/dt = v… (I)

(Considering motion in a straight line only)

And we also have the equation

dv/dt = a…(II)

Simply replacing the v in eqn (II) by eqn (I), we find

d2S/dt^2 = a…(III)

This is what we need to solve. It's easy.

You know,

d2S/dt^2 = d/dt(dS/dt) = a

⟹ dS/dt = ∫adt = at+c1

Since, dS/dt is the velocity of the particle,

Therefore, at t = 0, dS/dt|t = 0 = u

⟹ u = a∗0 + c1 = c1

⟹ c1 = u

Therefore, dS/dt = u + at

Thus, S = ∫(udt + atdt)

⟹ S = ut + 1/2at^2 +c^2

If say, the particle is already having a displacement S0 the moment you start measuring it's motion. Then, at t = 0, S = S0

This makes S = S0 +ut + 1/2at^2

Since, in most of the practical cases, we start measuring a motion when the particle starts displacing (i.e., when S0=0 ),

We get

S = ut + 1/2at^2

Hope it helps :)

5 0

3 years ago