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If a 15 kg mass accelerates at a rate of 4 m/s2, what net force acts on it?

jarptica [38.1K]

<h2>Answer: D 60N</h2>

<h3>Explanation:</h3>

Mass(M)=15 kg

Acceleration(A)=4 m/s2

Force=?

Now,

Force(F)=M×A

F=15×4

F=60N Ans

5 0

2 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

2 years ago

What will happen if you put a usb drive and a magnet too closely together?

kakasveta [241]

<span>If you put a magnet right next to a USB drive, depending on the strength of the magnet and the amount of steel, nickel or cobalt used in the construction of that particular model of USB drive, the drive would either adhere to, or not adhere to, the magnet. This would cause no other significant effects. The storage of data in solid state form (as in USB drives) is not magnetic in nature, so no deletion or any other damage of the stored data would occur.</span>

6 0

2 years ago

Read 2 more answers

Paola can flex her legs from a bent position through a distance of 20.1 cm. Paola leaves the ground when her legs are straight,

makkiz [27]

The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/ $s^{2}$ .

Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.

To determine Paola's acceleration, the third equation of free fall is appropriate.

i.e $V^{2}$ = $U^{2}$ ± 2as