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ikadub [295]
2 years ago
14

Can a vector have a component equal to zero and still have nonzero magnitude?

Physics
1 answer:
ExtremeBDS [4]2 years ago
8 0

Answer:

Yes

Explanation:

in a 2-D plane a vector can be resolved into its rectangular components. These are rectangular components because these are at right angles to each other. One component is directed along x-axis and is the termed as horizontal component and the other one is directed along y-axis and is termed as vertical component.

Sometimes, the vector is directed entirely along x-axis or y-axis. When the vector is along x-axis or parallel to x-axis all of its magnitude will be directed along horizontal component while its vertical component will be zero. Similarly, for a vector along y-axis, the horizontal component will be zero and the entire magnitude will be along y-axis.

Consider a vector V of magnitude 5 Newtons along x-axis. Since, V is along axis the angle made with the positive x-axis is 0.

The components of a vector are calculated as:

Horizontal component = V cos θ

Vertical component = V sin θ

θ is the angle made with positive x-axis in anti-clockwise direction which is 0 in this case.

So, the components of our vector will be:

Horizontal component = 5 cos(0) = 5

Vertical component = 5 sin(0) = 0

We have just showed, even if one of the components of a vector is 0, it still can have a non-zero magnitude.

The magnitude of a vector will be zero only when all of its components are equal to 0.

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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0
3 years ago
In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
worty [1.4K]

Answer:

in the direction of the applied force

Explanation:

8 0
3 years ago
the brakes on a car do 240000j of work in stopping the car. if the car travels a distance of 40m while the brakes re being appli
Tpy6a [65]
A joule is one Newton of force applied over a meter.
For every meter, the brakes put 240000N of force (N=Newtons).
For 40m, multiply the Newtons by 40.
240000N*40=9600000N
3 0
3 years ago
If a car travels 30 mi. north for 30 min., 60 mi. east for 1.0 hour, and 30 mi. south for 30 min., what is the average speed
Ainat [17]

Answer: 60mph

Explanation:

Given the following :

First leg travel:

Distance = 30 miles

Time of travel= 30 minutes = 0.5 hour

Second leg travel:

Distance = 60 miles

Time of travel = one hour

Average speed :

Speed = total Distance / time of travel

Total distance in miles = (30 + 60) miles = 90 miles

Total time of travel = 1 hour + 0.5 hour = 1.5 hours

Average speed = total distance traveled / total travel time

Average speed = 90 miles / 1.5 hours

Average speed = 60 miles / hour

= 60mph

3 0
3 years ago
Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is mo
kkurt [141]

Answer:increased

Explanation:

It is given that elevator speed is increasing while moving upward i.e.its acceleration is increasing .

This causes the apparent to be increased if measured using weighing machine.

considering upward direction to be positive

N-mg=ma

N=m(g+a)

where N=Normal reaction=Apparent weight

a=acceleration of Elevator

thus you feel as if your weight is increased.

6 0
3 years ago
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