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4 years ago

Draw the vector C⃗ =A⃗ +2B⃗ .

Physics

1 answer:

spin [16.1K]4 years ago

4 0

Answer:

A) C is 2 units towards right

B) C is 9 units towards right and 3 units towards up

C) C is 4 units towards right and 1 unit towards down

Explanation:

Drawings are given in the attachment.

A)Since,

A is 4 units towards right,

B is 1 unit towards left,

Then,

C is 2 units towards right.

B) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards left,

Then,

C is 9 units towards right and 3 units towards up.

C) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards right, and 1 unit towards down

Then,

C is 4 units towards right and 1 unit towards down.

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A Gaussian surface in the form of a hemisphere of radius R = 5.51 cm lies in a uniform electric field of magnitude E = 1.08 N/C.

Ksju [112]

Answer:

$Flux_{base}=-0.0103Nm^2/C$

$Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C$

Explanation:

a)At the base of the surface the Electric flux is easy to find, because the Electric Field is constant in magnitude and perpendicular to the flat surface:

$Flux_{base}=-E*S=-E*\pi*R^2=-1.08*\pi*0.0551^2=-0.0103Nm^2/C$

The Flux is negative because the Electric field goes into the surface

b) The Gaussian surface in form of a hemisphere encloses no net charge. The Gauss law says that the Flux of the electric field is proportional to the net charge enclosed. At this case, the charge is zero, then the total Flux is zero too.

$Flux_{total}=Flux_{base}+Flux_{curvePortion}=0$

Then:

$Flux_{curvePortion}=-Flux_{base}=0.0103Nm^2/C$

The Flux is positif because the Electric field goes out of the surface.

4 0

3 years ago

Use the work energy theorem to solve each of these problems and neglect air resistance in all cases. a) A branch falls from the

Sav [38]

Answer:

a)43.8 m/s

b)103.54 m/s

Explanation:

Work energy theorem

Δ $E_{k}=$ Δ $E_{g}$

$E_{k2} - E_{k1}=-(E_{g2} - E_{g1})\\\\ \frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2} - mgh_{1} = 0$

a) In this part v1=0 and h2=0

$\frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2}- mgh_{1} = 0\\\\\\\\\v_{2} =\sqrt{2gh_{1}} = \sqrt{2*9.8*98} =43.8 m/s$ =0

b) in this part v2=0, h1= 0, and h2= 545m

$\frac{1}{2} v^{2}_{1}= gh_{2} \\v_{1} =\sqrt{2gh_{2}}=103.54m/s$

5 0

3 years ago

atroni [7]

Answer:

ω = 0.1 rad/s

v = 0.002 m/s

Explanation:

The angular velcoity of the second hand of the clock can be found by:

ω = θ/t

where,

ω = Angular Speed

θ = Angular Displacement

t = time taken

Now, for one complete revolution of second hand of the clock:

θ = 2π rad

t = 60 s

Therefore,

ω = 2π rad/60 s

Now, for the linear speed (V):

V = rω

where,

V = Linear Speed of Second Hand = ?

r = radius = length of second hand = 0.02 m

Therefore,

V = (0.02 m)(0.1 rad/s)

5 0

3 years ago

Draw the vector C⃗ =A⃗ +2B⃗ .

spin [16.1K]

Answer:

A) C is 2 units towards right

B) C is 9 units towards right and 3 units towards up

C) C is 4 units towards right and 1 unit towards down

Explanation:

Drawings are given in the attachment.

A)Since,

A is 4 units towards right,

B is 1 unit towards left,

Then,

C is 2 units towards right.

B) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards left,

Then,

C is 9 units towards right and 3 units towards up.

C) Since,

A is 4 units towards right and 2 units towards up,

B is 1 unit towards right, and 1 unit towards down

Then,

C is 4 units towards right and 1 unit towards down.

4 0

4 years ago

A football wide receiver runs out from the line of scrimmage, turns around and runs 2 meters back toward the quarterback. Compar

nexus9112 [7]

Answer:

See below

Explanation:

TOTAL distance = x meters (not given how far he ran from scrimmage line)

Displacement = x - 2 meters ( although football uses yards not meters)

3 0

2 years ago