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1 year ago

3. The expression 0.62 x10^3 is equivalent to...

Physics

2 answers:

Korolek [52]1 year ago

5 0

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

Nitella [24]1 year ago

5 0

Answer:

620

Explanation:

0.62*1000=620

10^3= 10*10*10=1,000

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A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first bl

tatuchka [14]

By Newton's second law, assuming F is horizontal,

• the net horizontal force on the larger block is

F - µmg = 3mA

where µmg is the magnitude of friction felt by the larger block due to rubbing with the smaller one, µ is the coefficient of static friction between the two blocks, and A is the block's acceleration;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3mg is the weight of the larger block, and mg is the weight of the smaller block;

• the net horizontal force on the smaller block is

µmg = ma

where µmg is again the friction between the two blocks, but notice that this points in the same direction as F. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net vertical force on the smaller block is

mg - mg = 0

where mg is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for A and a :

A = (F - µmg) / (3m)

a = µg

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1 year ago

A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char

Aleksandr [31]

Answer:

Ek = 8,79 [J]

Explanation:

We are going to solve this problem, using the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁ = Ek + U₁

As charge are at rest Ek = 0

And U₁ has two components

U₁₂ = K * Q₁*Q₂ / 0,4 and U₃₂ = K*Q₃*Q₂ / 0,6

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4 ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6 ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J] ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then E₁ = E₁₂ + E₃₂

E₁ = 10,5 [J]

At the moment of Q₂ passing x = 40 cm or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂ = U₁₂ + U₃₂

U₁₂ = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7 ⇒ U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7 U₁₂ = 7,71 [J]

U₃₂ = - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3 ⇒ U₃₂ = - 9*20*10*10⁻³/0,3

U₃₂ = - 9*20*10⁻²/0,3

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as

E₂ = Ek + U₂ and E₂ = E₁

Then

Ek + U₂ = E₁

Ek = 10,5 - U₂

Ek = 10,5 - 1,71

Ek = 8,79 [J]

5 0

1 year ago

Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp

Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

v² = v₀² + 2 a x

0 = v₀² + 2 a x

a = - v₀² / 2 x

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

a/g = - (2v₀) 2 / 2xg

a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

a/g = (- v₀² / 2g) 2/x

a =2 (-v₀² / 2x₀g) g

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

a / g = - 20² / (2 0.05)

a / g = - 4000 / g

a / g = - 4000 / 9.8 = 408

a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

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2 years ago

A confused dragonfly flies forward and backward in a straight line. Its motion is shown on the following graph of horizontal pos

sertanlavr [38]

Answer:

Explanation:

Assuming your graph and question match the attachment, the average speed is 0. The bug ends up where it started, so its displacement is zero.

average speed = displacement/time = 0/(8 s)

average speed = 0

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2 years ago

Ben starts walking along a path at 3 3 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the s

Drupady [299]

Answer:

3 hours

Explanation:

Given:

- The speed of Ben v_b = 3 mi/h

- The speed of Amanda v_a = 6 mi/h

- The total time taken to cover distance(d) by ben = t_b

Find:

How long will it be before Amanda catches up to Ben?

Solution:

- The distance d traveled by Ben:

d = v_b*t_b

d = 3*t_b

- The distance d traveled by Amanda:

d = v_a*t_a

d = 6*t_a

- Equate the distance as when they meet:

3*t_b = 6*t_a

- Where ,

t_b = t_a + 1.5

t_a = t_b - 1.5

- Substitute the time relationship in distance relationship:

3*t_b = 6*(t_b - 1.5)

3*t_b = 6*1.5

t_b = 2*1.5 = 3 h

- Hence, It would take 3 hours since Ben starts walking that amanda catches up.

4 0

2 years ago