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3 years ago

Please answer this I'm sorry for annoying you

Physics

1 answer:

GalinKa [24]3 years ago

4 0

The resultant force points between up and right.

So if we're going to resolve it into two separate forces, they have to provide some force up and some force to the right.

The only pair of forces among the choices that does that is choice-A .

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What are the Si prefix values

Law Incorporation [45]

centic10-2millim10-3microu [footnote 2]10-6nanon10-9

4 0

3 years ago

calculate the currrent passing acrooss a bulb if it has a resistane of 15 ohms and a potentinal difference of 6v is applied acro

lawyer [7]

V = IR

6 V = I × 15 Ohms

I = 6 V : 15 Ohms

I = 0,4 A

#LearnWithEXO

4 0

3 years ago

A 5.93 kg ball is attached to the top of a vertical pole with a 2.35 m length of massless string. The ball is struck, causing it

Delvig [45]

Answer

given,

mass of ball = 5.93 kg

length of the string = 2.35 m

revolve with velocity of 4.75 m/s

acceleration due to gravity = 9.81 m/s²

T cos θ = mg

T cos θ = $5.93\times 9.81$

T cos θ = 58.17

$T sin \theta =\dfrac{mv^2}{r}$

$T sin \theta =\dfrac{5.93\times 4.75^2}{2.35 sin \theta}$

$T sin^2 \theta =56.93$

$sin^2 \theta = 1 - cos^2 \theta$

$T (1 - cos^2 \theta) =56.93$

$T (1 - (\dfrac{58.17}{T})^2) =56.93$

T² - 56.93T - 3383.75 = 0

T = 93.22 N

$cos \theta = \dfrac{58.17}{93.22}$

θ = 51.39°

6 0

4 years ago

A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on t

iris [78.8K]

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

5 0

3 years ago

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.320 with the floor. If

mart [117]

Answer: 29.50 m

Explanation: In order to calculate the higher accelation to stop a train without moving the crates inside the wagon which is traveling at constat speed we have to use the second Newton law so that:

f=μ*N the friction force is equal to coefficient of static friction multiply the normal force (m*g).

f=m.a=μ*N= m*a= μ*m*g= m*a

then

a=μ*g=0.32*9.8m/s^2= 3.14 m/s^2

With this value we can determine the short distance to stop the train

as follows:

x= vo*t- (a/2)* t^2

Vf=0= vo-a*t then t=vo/a

Finally; x=vo*vo/a-a/2*(vo/a)^2=vo^2/2a= (49*1000/3600)^2/(2*3.14)=29.50 m

5 0

3 years ago