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RoseWind [281]
3 years ago
13

Vector A has components Ax=1.30cm, Ay= 2.25cm; vector B has components Bx=4.10cm, By=-3.75cm.

Physics
1 answer:
geniusboy [140]3 years ago
6 0
We  are given with the x and y components of Vector A and B. In this case, we compute the resultant of both components of each vector. The vector is equal to the square root of the sum of the squares of the components. A is equal to 2.60 cm. B is equal to 5.56 cm. B is found in quadrant Iv and has an angle of 42.447 degrees as a terminal angle. A has an angle of 59.98 degrees. 
a. 5.6082 < -15.53 degreesc. 6.63 <-64.98 degreesb. x = 6.63 cos -64.98 degrees = 2.80   y = 6.63 sin -64.98 degrees = -6.00
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Objects 1 and 2 attract each other with a gravitational force of 12 units. If the mass of Object 2 is tripled, then the new grav
olasank [31]

Explanation:

Fgravity = G*(mass1*mass2)/D².

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between the objects.

so, if you triple one of the masses, what does that do to our equation ?

Fgravitynew = G*(3*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity

so, the right answer is 3×12 = 36 units.

5 0
2 years ago
Determine the projection (magnitude and sign), or component, of vector v1 along the direction of vector v2. Your answer could be
professor190 [17]

Answer:

- 1.07 ft

Explanation:

V1 = (-5, 7, 2)

V2 = (3, 1, 2)

Projection of v1 along v2, we use the following formula

=\frac{\overrightarrow{V1}.\overrightarrow{V2}}{V2}

So, the dot product of V1 and V2 is = - 5 (3) + 7 (1) + 2 (2) = -15 + 7 + 4 = -4

The magnitude of vector V2 is given by

= \sqrt{3^{2}+1^{2}+2^{2}}=3.74

So, the projection of V1 along V2 = - 4 / 3.74 = - 1.07 ft

Thus, the projection of V1 along V2 is - 1.07 ft.

so we need to find the direction of v2

7 0
3 years ago
Please help me solve this and give an explanation​
Arlecino [84]

Answer:

6.5

Explanation:

Because 1.5+5=6.5

7 0
2 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 4.57×10−21 n
hichkok12 [17]

Answer:

891 excess electrons must be present on each sphere

Explanation:

One Charge = q1 = q

Force = F = 4.57*10^-21 N  

Other charge = q2 =q

Distance = r = 20 cm = 0.2 m  

permittivity of free space = eo =8.854×10−12 C^2/ (N.m^2)  

Using Coulomb's law,

F=[1/4pieo]q1q2/r^2

F = [1/4pieo]q^2 / r^2

q^2 =F [4pieo]r^2

q =  r*sq rt F[4pieo]

 q=0.2* sq rt[ 4.57 x 10^-21]*[4*3.1416*8.854*10^-12]

q = 1.42614*10^ -16 C

number of electrons = n = q/e=1.42614*10^ -16 /1.6*10^-19

n =891

 891 excess electrons must be present on each sphere  

5 0
3 years ago
People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means th
fiasKO [112]

Answer:

636.619772368 A

Explanation:

\tau = Torque = 1\times 10^{-3}\ N/m

B = Magnetic field of Earth = 5\times 10^{-5}\ T

A = Area

d = Diameter = 20 cm

Current is given by

I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A

The current is 636.619772368 A

8 0
3 years ago
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