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xxTIMURxx [149]
3 years ago
7

two charges + 2.6 uc and -5.4uc experience an attractive force of 6.5mN. What is the separation between the charges?

Physics
1 answer:
djyliett [7]3 years ago
7 0

Answer:

...

Explanation:

....

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Science help please!
OLga [1]

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

3 0
3 years ago
I will Mark as the brainliest answer​
Kobotan [32]

Answer: looks good

Explanation:

3 0
3 years ago
Read 2 more answers
the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m
olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

x = \frac {-nmg}{k}

Substituting into the formula, we have;

x = \frac {-49*0.002*9.8}{24}

x = \frac {-0.9604}{24}

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0
3 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
In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in
cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

T=2\pi w

But we know as well that w is given by,

w=\sqrt{\frac{k}{m}}

Replacing,

w=\frac{2\pi}{T}= \sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{k}{m}}\\T=2\pi\sqrt{\frac{0.4}{23.1}}

So we have that

T=0.827s

7 0
3 years ago
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