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jolli1 [7]
3 years ago
9

Which of the following does not affect gas pressure

Physics
1 answer:
viktelen [127]3 years ago
4 0

Answer:

I NEED OPTIONS

Explanation:

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Joseph studied whether different materials can block certain electromagnetic waves by testing television reception in different
gizmo_the_mogwai [7]
Using the same antenna. Now he doesn't know if it was the antenna that caused the change in reception. so he wasn't only measuring the reception in his house he was measuring it based on different antennas
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3 years ago
A rectangular piece of​ cardboard, whose area is 352 square​ centimeters, is made into an open box by cutting a 2​-centimeter sq
Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

xy=352

y=\dfrac{352}{x} -------- (1)

The volume of the rectangular piece

2(x-4)(\dfrac{352}{x}-4)=432

x^2-38x+352=0

(x-16)(x-22)=0

x=16,22

Put the value of x in the equation (I)

For x = 16

y=\dfrac{352}{16}=22

For x = 22

y=\dfrac{352}{22}=16

Dimension of cardboard is 22 m by 16 m

8 0
3 years ago
What happens when you decrease the thrust on your scooter? A. You stop B. Nothing happens C. You fall over D. You speed up Reset
mars1129 [50]

Answer:

D. You speed up

Explanation:

hope it helps

4 0
3 years ago
You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C
jarptica [38.1K]

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

4 0
3 years ago
How does the strong nuclear force hold the nucleus of an atom together?
erastovalidia [21]

The strong nuclear force holds the nucleus of an atom together.  

Somehow, it overcomes the electrical force of repulsion between protons in the nucleus, which all have the same charge but still stay close together somehow. (b)

3 0
3 years ago
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