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3 years ago

A 75.0kg teacher (including the parachute) is skydiving! As the parachute opens, the system experiences a 1000N drag.

Physics

1 answer:

Ronch [10]3 years ago

3 0

Answer:

Net force = 250 N

Explanation:

Given that,

The mass of a teacher, m = 75 kg

As the parachute opens, the system experiences a 1000N drag.

We need to find the magnitude of net force acting on the teacher. 2 forces are acting on her i.e. drag force and weight.

Net force,

F = mg-D

= 75(10)-1000

= -250 N

Hence, the net force is 250 N and it is acting in the upward direction.

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The force of gravity is also known as

kolbaska11 [484]

Answer:

Gravitation

Explanation:

Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. ... On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth's mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects.

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3 years ago

Suppose Earth's mass increased but Earth's diame-

navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0

3 years ago

A police officer in hot pursuit of a criminal drives her car through an unbanked circular (horizontal) turn of radius 300 m at a

Mamont248 [21]

Answer:

The angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is <u>10°.</u>

Explanation:

Given:

Mass of the driver is, $m=55\ kg$

Radius of circular turn is, $R=300\ m$

Linear speed of the car is, $v=22.2\ m/s$

Since, the car makes a circular turn, the driver experiences a centripetal force radially inward towards the center of the circular turn. Also, the driver experiences a downward force due to her weight. Therefore, two forces act on the driver which are at right angles to each other.

The forces are:

1. Weight = $mg=55\times 9.8=539\ N$

2. Centripetal force, 'F', which is given as:

$F=\frac{mv^2}{R}\\F=\frac{55\times (22.2)^2}{300}\\\\F=\frac{55\times 492.84}{300}\\\\F=\frac{27106.2}{300}=90.354\ N$

Now, the angle of the net force acting on the driver with respect to the vertical is given by the tan ratio of the centripetal force (Horizontal force) and the weight (Vertical force) and is shown in the triangle below. Thus,

$\tan \theta=\frac{90.354}{539}\\\tan \theta=0.1676\\\theta=\tan^{-1}(0.1676)=9.52\approx 10$ °

Therefore, the angle (relative to vertical) of the net force of the car seat on the officer to the nearest degree is 10°.

3 0

2 years ago

Write relationship between hertz and megahertz

dusya [7]

Explanation:

1 mega Hertz = 1000000 hertz

8 0

2 years ago

A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di

siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

$E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\$

part a)

Electric Field strength at point P: a = 0.75 m

$E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C$

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

$E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o$

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

6 0

3 years ago