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3 years ago

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The graph shows how the vertical velocity of a parachutist changes from the moment the parachutist jumps from the aircraft until

landing on the ground. Using the idea of forces, explain why the parachutist reaches a terminal velocity and why opening the parachute reduces the terminal velocity.

1 answer:

soldi70 [24.7K]3 years ago

4 0

Answer:

BUT WHY

Explanation:

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A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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3 0

1 year ago

I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving

snow_lady [41]

Answer:

$9.7\times 10^{-4}\ rad/s$

Explanation:

Given:

$m=500 kg\\F=3000 N$

$Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\ know\ that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}$

= $\frac{6}{6371 \times 10^3\ m}$

= $9.7\times 10^{-4}\ rad/s$

7 0

3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

2 years ago

ONE EASY QUESTION

Brrunno [24]

Answer:

its tension force which acts in a string

Explanation:

need a thanks and thats it

8 0

3 years ago

Rita throws a ball straight up into the air and catches it at the

Kaylis [27]

(1) The potential energy at the top of the ball’s motion is 18 J.

(3) The kinetic energy increases as the potential energy decreases.

(4) The kinetic energy decreases as the potential energy increases.

(5) The total mechanical energy of the ball stays constant.

Explanation:

The total mechanical energy of the ball is equal to the sum of its kinetic energy (K, energy due to the motion) and its potential energy (U, energy due to the height of the ball). Mathematically:

$E=K+U$

In absence of friction, the mechanical energy of the ball is conserved, so in this case, it is always equal to 18 J. Let's now use this information to analyze each of the given statements:

(1) The potential energy at the top of the ball’s motion is 18 J. --> TRUE. In fact, at the top, the ball's speed becomes zero, so its kinetic energy is zero: K = 0. This means that all the mechanical energy of the ball is potential energy, therefore

E = U = 18 J

(2) The kinetic energy is less when the ball is thrown than when it is caught. --> FALSE. As we said, in absence of friction, the mechanical energy is conserved, therefore it always remains equal to 18 J.

(3) The kinetic energy increases as the potential energy decreases. --> TRUE. As we said, the sum of potential+kinetic energy remains constant:

E = K + U = 18 J

therefore, when the potential energy decreases, the kinetic energy increases.

(4)The kinetic energy decreases as the potential energy increases. --> TRUE. For the same reason described in (3).

(5)The total mechanical energy of the ball stays constant. --> TRUE. As we said at the beginning, the total mechanical energy is constant.

(6) The mechanical energy decreases as the ball moves up and increases as the ball comes down. --> FALSE. As we said, the mechanical energy remains constant, so it cannot change.

Learn more about kinetic and potential energy:

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6 0

3 years ago

Read 2 more answers