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4 years ago

Do you guys know any videos that’ll help understand soh cah toa

1 answer:

7 0

The video called “SOH CAH TOA: how to solve a right triangle.”
This is explained very well. Make sure you write down the “equations” of SOH CAH and TOA. It helps a lot to be able to go back and plug in from the equation.

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What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed

tekilochka [14]

Answer:

charge will be equal to $2.03\times 10^{-5}C$

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity $g=9.8m/sec^2$

Electric field E = 700 N/C

Electric force will be equal to $F=qE$ , here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

$q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C$

So charge will be equal to $2.03\times 10^{-5}C$

8 0

3 years ago

A spring of spring constant k is attached to a support at the bottom of a ramp that makes an angle θ with the horizontal. A bloc

Nikitich [7]

Answer:

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

Explanation:

From the law of conservation of energy

Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr

$0.5kd^{2}=0.5mv^{2}+mgLsin\theta+\mu_{k}mgcos\theta x$

$x(mgsin\theta+\mu_{k}mgcos\theta)=0.5kd^{2}$

$x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

The required distance from A to B is $x=\frac {kd^{2}}{2(mgsin\theta +\mu_{k}mgcos\theta)}$

5 0

3 years ago

An object traveling at a constant speed but with a changing direction is accelerating.

prohojiy [21]

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up". It means ANY change in
the speed or direction of motion. So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

6 0

3 years ago

Read 2 more answers

A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000

34kurt

Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N

4 0

3 years ago

A 103 kg physics professor has fallen into the Grand Canyon. Luckily, he managed to grab a branch and is now hanging 93 m below

siniylev [52]

Answer:

125.83672 seconds

Explanation:

P = Power of the horse = 1 hp = 746 W (as it is not given we have assumed the horse has the power of 1 hp)

m = Mass of professor = 103 kg

g = Acceleration due to gravity = 9.8 m/s²

h = Height of professor = 93 m

Work done would be equal to the potential energy

$W=mgh\\\Rightarrow W=103\times 9.8\times 93\\\Rightarrow W=93874.2\ J$

Power is given by

$P=\frac{W}{t}\\\Rightarrow t=\frac{W}{P}\\\Rightarrow t=\frac{93874.2}{746}\\\Rightarrow t=125.83672\ seconds$

The time taken by the horse to pull the professor is 125.83672 seconds

6 0

3 years ago