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2 years ago

1 lb equals how many grams

Physics

2 answers:

shusha [124]2 years ago

8 0

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

Amiraneli [1.4K]2 years ago

8 0

453.592 grams hope this helps!

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A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

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2 years ago

Isabella deja caer accidentalmente un bolígrafo desde su balcón mientras celebra que resolvió satisfactoriamente un problema de

jok3333 [9.3K]

Answer:

i cant read spanish lol

Explanation:

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2 years ago

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$

$v_{i}$ is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

$\Delta x = 5(6) - \frac{4}{2}.6^{2}$

$\Delta x = 30 - 2.36$

$\Delta x$ = - 42

Displacement of the boat for t=6.0s interval is $\Delta x$ = - 42m, i.e., 42 m to the left.

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2 years ago

A lightning bolt transfers 6.0 coulombs of charge from a cloud to the ground in 2.0 x 10-3 second. what is the average current d

AlladinOne [14]

The current is defined as the amount of charge transferred through a certain point in a certain time interval:
$I= \frac{Q}{\Delta t}$
where
I is the current
Q is the charge
$\Delta t$ is the time interval

For the lightning bolt in our problem, Q=6.0 C and $\Delta t= 2.0 \cdot 10^{-3}s$ , so the average current during the event is
$I= \frac{Q}{\Delta t} = \frac{6.0 C}{2.0 \cdot 10^{-3} s}=3000 A$

4 0

2 years ago

Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4

weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530 P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= - 1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny= -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

$F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2} }$

$F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2} }$

Fe= 2579.68 P

Calculation of the direction of equilibrant force (α)

$\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )$

$\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )$

α= 24.8°

Look at the attached graphic

6 0

2 years ago