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Daniel [21]
2 years ago
3

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0 ∘ below the positive x axis. A second vector, B⃗ , has a ma

gnitude of 75.0 m and points in a direction 50.0 ∘ above the positive x axis.
a) Sketch the vectors A⃗ , B⃗ , and C⃗=A⃗+B⃗ .

b) Using the component method of vector addition, find the magnitude of the vector C⃗ .

c) Using the component method of vector addition, find the direction of the vector C

Physics
1 answer:
AysviL [449]2 years ago
4 0

Answer:

Explanation:

Check attachment for solution

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Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l
s2008m [1.1K]
The magnetic field strength in a coil is directly proportional to the number of turns, or loops, in the coil.
Therefore, when there are four loops instead of one, the magnetic field strength has increased four times, making it harder to push the magnet in.
6 0
2 years ago
Read 2 more answers
The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
A spring with a spring constant of 2500 n/m. is stretched 4.00 cm. what is the work required to stretch the spring?
Yuri [45]
W = 1/2k*x^2.

k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).

W = 1/2(2500)(0.04)^2 = 2J.
5 0
2 years ago
Read 2 more answers
A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

Given

mass of bullet m=5 gm

mass of wood block M=1 kg

Length of string L=2 m

Center of mass rises to an height of 0.38 cm

initial velocity of bullet u=450 m/s

let v_1 and v_2 be the velocity of bullet and block after collision

Conserving momentum

mu=mv_1+Mv_2 -------------1

Now after the collision block rises to an height of 0.38 cm

Conserving Energy for block

kinetic energy of block at bottom=Gain in Potential Energy

\frac{Mv_2^2}{2}=Mgh_{cm}

v_2=\sqrt{2gh_{cm}}

v_2=\sqrt{2\times 9.8\times 0.38}

v_2=0.272 m/s

substitute the value of v_2 in equation 1

5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exert
Dimas [21]

A falling skydiver opens his parachute. A short time later, the weight of the skydiver-parachute system and the drag force exerted on the system are equal in magnitude. The following statements predicts the motion of the skydiver at this time

<u>The skydiver is moving downward with constant speed.</u>

Explanation:

Immediately on leaving the aircraft, the skydiver accelerates downwards due to the force of gravity. There is no air resistance acting in the upwards direction, and there is a resultant force acting downwards. The skydiver accelerates towards the ground.

The forces acting on a falling leaf are : gravity and air resistance.

The net force and the acceleration on the falling skydiver is upward.

An upward net force on a downward falling object would cause that object to slow down. The skydiver thus slows down.

As the speed decreases, the amount of air resistance also decreases until once more the skydiver reaches a terminal velocity.

<u>A skydiver falling at a constant speed opens his parachute. When the skydiver is falling, the forces are unbalanced.</u>

6 0
2 years ago
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