Two 8.0 Ω lightbulbs are connected in a 12 V parallel circuit. What is the power of both glowing bulbs?

) is it possible for one component of a vector to be zero, while the vector itself is not zero?

Natalka [10]

The least number of component of a vector quantity is two. These are the x-component and the y-component.

The resultant vector, or vector as we refer to it in this item, can be calculated through the equation,
RV = sqrt ((Vx)² + (Vy)²)

From the equation, it can be noted that if we let Vx equal to zero,
RV = Vy

Similarly, if we let Vy be equal to zero then,
RV = Vx

Thus, it is still possible for the vector to become nonzero even if one of its components is zero.

8 0

3 years ago

A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th

melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore, the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F = 624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0

3 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. a

USPshnik [31]

Refer to the diagram shown below.

Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force

The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m

By definition,
T = I*α

Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²

Answer: 105.93 kg-m²

6 0

2 years ago

(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the per

Blizzard [7]

Answer:

$\huge\boxed{T = 0.025\ seconds}$

Explanation:

<u>Given:</u>

Frequency = f = 40 Hz

<u>Required:</u>

Time period = T = ?

<u>Formula:</u>

$\sf T = 1 / f$

<u>Solution:</u>

T = 1 / 40

T = 0.025 seconds

$\rule[225]{225}{2}$

Hope this helped!

4 0

3 years ago

What does 760mmHg represents??

malfutka [58]

Explanation:

760 mmHg (millimeters of mercury) is a measure of atmospheric pressure. It represents the height of a column of mercury at which the static pressure at the bottom is equal to the atmospheric pressure.

1 atm = 760 mmHg = 101,325 Pa = 14.7 psi

8 0

3 years ago