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3 years ago

During the final stages of descent, asky diver with an

Physics

1 answer:

OLEGan [10]3 years ago

3 0

Answer:

Gravity: downwards

Air drag and air-pressure on the inner surface of the the parachute: Upwards

Explanation:

If a sky-diver is in the final stages of his descend with open parachute such that the wind is calm and it does not blows him laterally.

In such a condition the air resistance in the form of drag and the pressure force due to the air captured in the parachute are acting in the upward direction which balance the force of gravity on the body. But this situation may occur momentarily and then again the diver must begin to slowly descend.

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A yellow train of mass 100 kg is moving at 8 m/s towards an orange train of mass 200 kg traveling on the opposite direction on t

vladimir1956 [14]

Mass of yellow train, my = 100 kg

Initial Velocity of yellow train, = 8 m/s

mass of orange train = 200 kg

Initial Velocity of orange train = -1 m/s (since it moves opposite direction to the yellow train, we will put negative to show the opposite direction)

To calculate the initial momentum of both trains, we will use the principle of conservation of momentum which

The sum of initial momentum = the sum of final momentum

Since the question only wants the sum of initial momentum,

(100)(8) + (200)(-1) = 600 m/s

8 0

1 year ago

Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre

kari74 [83]

The answers to your questions are as written below:

The objects that represents a negatively charged particle is : Image B
The object that represents a positively charged molecule is : Image A
The object that represents an uncharged molecule is : Image C
The object the will not move when in an electric fied is : Image C

<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

#SPJ4

attached below is the missing image

8 0

1 year ago

WILL MARK BRAINLIST. HELP ASAP

Pachacha [2.7K]

Answer:

Less than the distance

3 0

2 years ago

A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

Blababa [14]

Answer:

The maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

$\tau=NIAB\ \sin\theta$

For maximum torque, $\theta=90^{\circ}$

$\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m$

So, the maximum torque in the coil is $4.9\times 10^{-3}\ N-m$ .

4 0

2 years ago

An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a

Morgarella [4.7K]

Answer:

0.37 m

Explanation:

The angular frequency, ω, of a loaded spring is related to the period, T, by

$\omega = \dfrac{2\pi}{T}$

The maximum velocity of the oscillation occurs at the equilibrium point and is given by

$v = \omega A$

A is the amplitude or maximum displacement from the equilibrium.

$v = \dfrac{2\pi A}{T}$

From the the question, T = 0.58 and A = 25 cm = 0.25 m. Taking π as 3.142,

$v = \dfrac{2\times3.142\times0.25\text{ m}}{0.58\text{ s}} = 2.71 \text{ m/s}$

To determine the height we reached, we consider the beginning of the vertical motion as the equilibrium point with velocity, v. Since it is against gravity, acceleration of gravity is negative. At maximum height, the final velocity is 0 m/s. We use the equation

$v_f^2 = v_i^2+2ah$

$v_f$ is the final velocity, $v_i$ is the initial velocity (same as v above), a is acceleration of gravity and h is the height.

$h = \dfrac{v_f^2 - v_i^2}{2a}$

$h = \dfrac{0^2 - 2.71^2}{2\times-9.81} = 0.37 \text{ m}$

3 0

2 years ago