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3 years ago

What is a non example of atmosphere

Physics

2 answers:

Shkiper50 [21]3 years ago

6 0

Google it and I bet you find the correct answer

Oxana [17]3 years ago

4 0

A stuffed animal is a non example of atmosphere.

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4. How does the type of medium affect a sound wave?

jek_recluse [69]

Answer:

The type of medium affects a sound wave as sound travels with the help of the vibration in particles.

Explanation:

As different mediums have different amount and size of particles, for example, the speed of sound is faster through solid than liquid as solids have closely packed particles whereas liquids are loosely packed.

5 0

2 years ago

The image produced by a convex mirror is always closer to the mirror than it would be in a plane mirror for the same object dist

rosijanka [135]

Answer:

True

Explanation:

The image produced a convex mirror is always virtual irrespective of location. The size of the image is always smaller than the object. In a plane mirror the distance of the object and the distance of the image is same. But in a convex the image distance is always less than the object distance.

So, this statement is true.

6 0

3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago

A bald eagle is flying to the left with a speed of 34 meters

Shtirlitz [24]

Answer:

the speed after 3 seconds is 10 m/s

Explanation:

The computation of the speed is shown below:

As we know that

V = U + at

Here,

U = 34 m/s

a = - 8 m/s²

t = 3 Sec

V = velocity after 3 sec

V = 34 + (-8)3

= 34 - 24

V = 10 m/s

Hence, the speed after 3 seconds is 10 m/s

4 0

2 years ago

Explain how to find the angle between two nonzero vectors. Choose the correct answer below. A. The angle between two nonzero vec

kozerog [31]

Answer:

θ = Cos⁻¹[A.B/|A||B|]

A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result

Explanation:

We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:

A.B = |A||B| Cosθ

where,

A = 1st Non-Zero Vector

B = 2nd Non-Zero Vector

|A| = Magnitude of Vector A

|B| = Magnitude of Vector B

θ = Angle between vector A and B

Therefore,

Cos θ = A.B/|A||B|

Hence, the correct answer will be:

<u>A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result</u>

3 0

2 years ago