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SashulF [63]
3 years ago
8

PLEASE HELP WILL GIVE BRANIST IT WOULD REALLY BE APPRICATIED

Physics
1 answer:
Natasha_Volkova [10]3 years ago
4 0

Answer:

solids are strong and compact, they are not compressable.

liquids are flexible and less compacts they can not be coompresed

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Is the relationship between the length of a pendulum and its period is valid at all times?
Stels [109]
Yes it is valid all the times under the consideration of acceleration due to gravity .it is not valid on space where there is no influence of gravity
4 0
3 years ago
A driver notices an upcoming speed limit change from 45 mi/h (20 m/s) to 25 mi/h (11 m/s). If she estimates
zloy xaker [14]

Answer:

-2.79 m/s²

Explanation:

Given:

v₀ = 20 m/s

v = 11 m/s

Δx = 50 m

Find: a

v² = v₀² + 2aΔx

(11 m/s)² = (20 m/s)² + 2a (50 m)

a = -2.79 m/s²

Round as needed.

8 0
4 years ago
Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4
Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0
3 years ago
An object moving forward at 10 m/s suddenly begins moving backward at 5 m/s. What is the most accurate explanation for this chan
Vinil7 [7]
My best guess would be C.
8 0
3 years ago
Read 2 more answers
PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!
drek231 [11]
I would say true. If you are calculating using vectors than it would need both...
4 0
3 years ago
Read 2 more answers
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