Why does the arrow labeled F norm point upward?

Calculate the time period of simple pendulum whose length is 98.2cm

alekssr [168]

The time period resulting in oscillations will be 1.986 seconds.

<h3>What is the period of oscillation?</h3>

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The time period of the oscillation is;

$\rm T = 2 \pi\sqrt{\frac{L}{g}} \\\\ \rm T = 2 \times 3.14 \times \sqrt{ \frac{98.2 \ \times 10^{-2} \ m}{9.81 \ m/s^2}} \\\\ T= 1.986 \ sec$

Hence the time period resulting oscillations will be 1.986 seconds.

To learn more about the time period of oscillation refer to the link;

brainly.com/question/20070798

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5 0

1 year ago

A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the

butalik [34]

The complete question is:

Study the scenario. A firework is ignited, and explodes with a flash and a loud bang as it is blown apart. The system consists of: the firework, the earth, and the air. which choice best describes how energy is transformed in the system?

A) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. the energy is in several forms including sound and light, and mechanical energy of the fragments of the firecracker that are launched through the air. Eventually all the energy released is transformed into thermal energy.

B) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into mechanical energy.

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

D) When the firework is ignited, a chemical reaction absorbs energy from the surrounding environment. The energy is taken in in several forms including sound and light, and mechanical energy of the fragments being launched through the air. Eventually all the energy is transformed into thermal energy.

Answer:

C) When the firework is ignited, a chemical reaction releases energy in several forms, including sound, light, and the mechanical energy of the fragments being launched through the air. Eventually all the energy released is transformed into thermal or mechanical energy.

Explanation:

Energy is released from the system, not absorbed.

All the sound, light and movement of the debris is as a result of energy transformation from the chemical energy.

Eventually, most of the energy are finally wasted away as heat energy.

Some of the energy is used up by the flying particles from the fireworks.

8 0

2 years ago

An accelerating voltage of 2.47 x 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horiz

Dmitry [639]

Answer:

6.3445×10⁻¹⁶ m

Explanation:

E = Accelerating voltage = 2.47×10³ V

m = Mass of electron

Distance electron travels = 33.5 cm = 0.335 cm

$E=\frac{mv^2}{2}\\\Rightarrow v=\sqrt{\frac{2E}{m}}\\\Rightarrow v=\sqrt{\frac{2\times 2470\times 1.6\times 10^{-19}}{9.11\times 10^{-31}}}\\\Rightarrow v=29455356.08671\ m/s$

Deflection by Earth's Gravity

$\Delta =\frac {gt^2}{2}$

Now, Time = Distance/Velocity

$\Delta =\frac {g\frac{s^2}{v^2}}{2}\\\Rightarrow \Delta =\frac{9.81\frac{0.335^2}{29455356.08671^2}}{2}\\\Rightarrow \Delta =6.3445\times 10^{-16}\ m$

∴ Magnitude of the deflection on the screen caused by the Earth's gravitational field is 6.3445×10⁻¹⁶ m

3 0

2 years ago

you are piloting a small plane and you want to reach an airport 450 km due south in 3.0 h a wind is blowing from the west 50.0 k

alex41 [277]

Answer:

You should choose airspeed 158.11 km/h at 18.4° west of south

Explanation:

The distance to the air port is 450 km due to south

You should to reach the airport in 3 hours

→ Velocity = distance ÷ time

→ Distance = 450 km , time = 3 hours

→ The velocity of your plane = 450 ÷ 3 = 150 km/h due to south

A wind is blowing from west 50 km/h

We need to know what heading and airspeed you should choose to

reach your destination

At first we must find the resultant velocity of your plane and the wind

The south and west are perpendicular, then the resultant velocity is

→ $v_{R}=\sqrt{(v_{p})^{2}+(v_{w})^{2}}$

→ $v_{p}=150$ km/h , $v_{w}=50$ km/h

→ $v_{R}=\sqrt{(150)^{2}+(50)^{2}}=158.11$ km/h

To cancel the velocity of the wind, the pilot should maintain the velocity

of the plane at 158.11 km/h

The direction of the velocity is the angle between the resultant velocity

and the vertical (south)

→ The direction of the velocity is $tan^{-1}\frac{50}{150}=18.4$ °

The direction of the velocity is 18.4° west of south

<em>You should choose airspeed 158.11 km/h at 18.4° west of south</em>

8 0

2 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

2 years ago