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tiny-mole [99]
3 years ago
5

25 PTs

Physics
1 answer:
Nuetrik [128]3 years ago
8 0
Answer:
They are both wrong!

Liquid oxygen really is a pale blue color.
I’ve seen it.

And they cant say that liquid and solid oxygen is blue which makes the sky blue because they’re not and it doesn’t make up for the color of the sky.

The sky is actually blue because It reflects more light than It can absorb
aka Rayleigh Scattering.

-HOPE THAT HELPED
You might be interested in
Block 1, of mass m₁ = 1.30 kg , moves along a frictionless air track with speed v₁ = 29.0 m/s. It collides with block 2, of mass
Alecsey [184]

Answer:

a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J

Explanation:

a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s

The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s

So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s

b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.

p₂ = (1.3 + 39.0)v = 40.3v

From the principle of conservation of momentum,

p₁ = p₂

37.7 kgm/s = 40.3v

v = 37.7/40.3 = 0.94 m/s

So the final velocity of the two-block system is 0.94 m/s

c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²

So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J

7 0
3 years ago
A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl
Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon m_1 and the mass of the basket m_2, then according to newton's second law:

(1). \:F = (m_1+m_2)a,

where a =0.265m/s^2 is the upward acceleration, and F = 688N is the net propelling force (counts the gravitational force).

Also, the tension T in the rope is 79.8 N more than the basket's weight; therefore,

(2). \:T = m_2g+79.8

and this tension must equal

T -m_2g =m_2a

(3). \:T = m_2g +m_2a

Combining equations (2) and (3) we get:

m_2a = 79.8

since a =0.265m/s^2, we have

\boxed{m_2 = 301.13kg}

Putting this into equation (1) and substituting the numerical values of F and a, we get:

688N = (m_1+301.13kg)(0.265m/s^2)

\boxed{m_1 = 2295 kg}

Thus, the mass of the balloon and the basket is  2295 kg and 301 kg respectively.

8 0
3 years ago
Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
3 years ago
A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
SVETLANKA909090 [29]

A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of x_0 = 5.0 m

- at t = 0, the initial velocity of the motorcyclist is v_0 = 15.0 m east

- The acceleration of the motorcyclist is constant and it is a=4.0 m/s^2

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

x(t)=x_0 + v_0t + \frac{1}{2}at^2

where t is the time.

Substituting t = 2.0 s, we find the position:

x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

v(t)=x'(t)=v_0 + at

where:

v_0=15.0 m/s is the initial velocity

a=4.0 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s

And therefore, the position at t = 2.5 s is:

x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0
3 years ago
Emboldened by the success of their late night keg pull in Exercise 61 above, our intrepid young scholars have decided to pay hom
alexdok [17]

Answer is answer

XD                                ssssss

5 0
3 years ago
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