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Lorico [155]
2 years ago
7

HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics
1 answer:
Diano4ka-milaya [45]2 years ago
6 0
B, it has a drop, so it is going to increase in speed
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An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going
guapka [62]
Negative acceleration d. I think
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3 years ago
Which ONE of the following pairs of physical quantities consists of one scalar and one vector quantity?
lyudmila [28]

Answer:

speed and acceleration

Explanation:

speed is a scalar quantity

acceleration is a vector quantity

7 0
2 years ago
A Jeep Rubicon has a mass of 2,630 kg. With a momentum of 55,200 kg m/s. What was the speed, in m/s, of the vehicle?
Annette [7]

Answer:

The speed of the vehicle was 21 m/s

Explanation:

Momentum can be defined as mass in motion

Momentum depends on the variables mass and velocity

The momentum of an object is equal to the mass of the object times the

velocity of the object

The given is:

→ A Jeep Rubicon has a mass of 2,630 kg

→ With a momentum of 55,200 kg m/s

We need to find the speed of the vehicle

→ The momentum = mass × speed

Substitute the values above in the rule

→ 55,200 = 2,630 × speed

Divide both sides by 2,630

→ speed = 20.99 ≅ 21 m/s

<em>The speed of the vehicle was 21 m/s</em>

4 0
3 years ago
A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be
Komok [63]

Answer:

Tissues that are damaged or injured.

Explanation:

Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.

Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.

4 0
3 years ago
You have a spring that stretches 0.070 m when a 0.10-kg block is attached to and hangs from it at position y0. Imagine that you
olga nikolaevna [1]

Answer:

a) Δy = 0.144 m

b) W = 0.145 J

c) Us = 0.32 J

d) ymax = 0.144 m

Explanation:

a) First let's find the spring constant using Hooke's Law

F = k*Δy   ⇒  k = F/Δy

where

F = m*g = 0.1 kg*9.81 m/s² = 0.981 N

and  Δy = 0.07 m. Hence

k = 0.981 N/0.07 m = 14.014 N/m ≈ 14 N/m

In order to find the position of the block when we let it go, we need to find the force that caused this expansion in the spring, we know that the reading of the scale was 3 N and this reading includes the force we want to find and the weight of the block, therefore:

f = 3 N - F = 3 N - 0.981 N = 2.019 N

Now that we have found the force we can use Hooke's Law in order to find the position of the block

f = k*Δy   ⇒   Δy = f/k

⇒   Δy = 2.019 N/14 N/m

⇒   Δy = 0.144 m

b) First, notice that there are two kind of potential energy: the potential energy in the spring and the potential energy due to the gravitational field:

W = ΔU

W = ΔUs + ΔUg

W = (Usf - Usi) + (Ugf - Ugi)

Notice that

Us = 0.5*k*y²

where

yf = 0.07 m + 0.144 m = 0.214 m  and

yi = 0.07 m

and we will take the zero level to be the equilibrium position where the block was hanging at rest. Hence

W = 0.5*k*(yf² - yi²) + m*g*(0 - Δy)

⇒ W = 0.5*14 N/m*((0.214 m)² - (0.07 m)²) + (0.1 kg)*(9.81 m/s²)*(0 - 0.144 m)

⇒ W = 0.145 J

c) When we let the block go the spring was stretched by

y = 0.07 m + 0.144 m = 0.214 m

Therefore:

Us = 0.5*k*y²

⇒ Us = 0.5*14 N/m*(0.214 m)²

⇒ Us = 0.32 J

d) Because the position that we pulled the block to it is considered as the amplitude for the vibrational motion that will happen after we release the block, then the maximum height the particle will reach above the equilibrium position is

ymax = Δy = 0.144 m

 

3 0
3 years ago
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