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Strike441 [17]
3 years ago
11

Vector A has a magnitude of 50 units and points in the positive x direction. A second vector, B , has a magnitude of 120 units a

nd points at an angle of 70 degrees below the x axis.
Part A

Which vector has the greater x component.

Part B

Which vector has the greater y component?
Physics
1 answer:
Alex Ar [27]3 years ago
8 0

A) Vector A

The x-component of a vector can be found by using the formula

v_x = v cos \theta

where

v is the magnitude of the vector

\theta is the angle between the x-axis and the direction of the vector

- Vector A has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its x-component is

A_x = A cos \theta_A = (50) cos 0^{\circ}=50

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ} (negative since it is below the x-axis), so the x-component is

B_x = B cos \theta_B = (120) cos (-70^{\circ})=41

So, vector A has the greater x component.

B) Vector B

Instead, the y-component of a vector can be found by using the formula

v_y = v sin \theta

Here we have

- Vector B has a magnitude of 50 units along the positive x-direction, so \theta_A = 0^{\circ}. So its y-component is

A_y = A sin \theta_A = (50) sin 0^{\circ}=0

- Vector B has a magnitude of 120 units and the direction is \theta_B = -70^{\circ}, so the y-component is

B_y = B sin \theta_B = (120) sin (-70^{\circ})=-112.7

where the negative sign means the direction is along negative y:

So, vector B has the greater y component.

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An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery
attashe74 [19]

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

Where ϵ0 is the permittivity of free space.  

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

        C1=K(ϵ0A/d)

        C1=KC   ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.  

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2  is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q   =  Q1

CV = C1V1

  CV = C1V1  -------2

We derived C1=KC from equation 1. Inserting this into equation 2

   CV = KCV1

 V1 = (CV)/KC

         V/K

       = 21/2.2

      = 9.5

4 0
3 years ago
A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit
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Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       F = \frac{\Delta p}{\Delta t}

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N

       ⇒ Fsk = 47.6 N (normal to the wall)

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Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

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A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

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The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

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The distance that the shuttle has traveled during the given time is  1820.7 Km.

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