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3 years ago

Why don't protons stick together?

Physics

2 answers:

professor190 [17]3 years ago

6 0

Answer:

The protons must feel a repulsive force from the other neighboring protons. This is where the strong nuclear force comes in. ... If they can't get that close, the strong force is too weak to make them stick together, and other competing forces (usually the electromagnetic force) can influence the particles to move apart.^o^

Alja [10]3 years ago

5 0

Answer:

Because, they are of the same charge that is, + and +, + can't go with another +.

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Zoe is setting up a track for a toy car. The track has a ramp that is 32° above horizontal. If Zoe wants the car to travel as a

jolli1 [7]

Answer:

Explanation:

Not enough information.

IF we ASSUME she wants the car to be at LAUNCH LEVEL after 1 second of flight.

THEN

The highest point will have zero vertical velocity and will have taken ½ second to get there. This means that the initial vertical velocity was

v = gt

vy₀ = 9.8(0.5)

vy₀ = 4.9 m/s

vsinθ = vy₀

v = vy₀/sinθ

v = 4.9/sin32

v = 9.2466...

v = 9.2 m/s

8 0

2 years ago

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm.

jasenka [17]

Answer:

$P_{1}-P_{2}=12675Pa$

Explanation:

From the equation of continuity we know that:

$v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s\\$

From Bernoulli equation we know that:

$P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}\\$

Now assuming $y_{1}=y_{2}$

$P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2} )\\ P_{1}-P_{2}=12675Pa$

5 0

3 years ago

A car with a mass of 800 g and velocity of 15 m/s collided with a truck of a velocity of 20 m/s, if the momentum is conserved wh

nydimaria [60]

Question: A car with a mass of 800 g and velocity of 15 m/s collided with a truck moving in opposite direction with a velocity of 20 m/s, if the momentum is conserved and they both move with a common velocity of 10 m/s, what is the mass of the truck?

Answer:

0.133 kg

Explanation:

Applying the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')................... Equation 1

Where m = mass of the car, m' = mass of the truck, u = initial velocity of the car, u' = initial velocity of the truck, V = common velocity.

From the question,

Given: m = 800 g = 0.8 kg, u = 15 m/s, u' = -20 m/s, V = 10 m/s

Substitute these values into equation 2

(0.8*15)+(m'*20) = 10(0.8+m')

Solve for m'

12-20m' = 8+10m'

-20m'-10m' = 8-12

-30m' = -4

m' = -4/-30

m' = 0.133 kg

4 0

3 years ago

BRAINLIEST. Agraph is probelow. The graph shows the speed of a car traveling east over a 12 second period. Based on the informat

gavmur [86]

Answer: Speeding up constantly

Explanation:

3 0

4 years ago

Read 2 more answers

Suppose a truck has a momentum of 40,120 kg • and a mass of 1,180 kg. what is the truck’s velocity?

klasskru [66]

The momentum of an object is the product between the mass m of the object and its velocity:
$p=mv$
The truck in our problem has momentum equal to
$p=40120 kg m/s$
and a mass of
$m=1180 kg$
so, if we re-arrange the previous formula, we can use these data to find the velocity of the truck:
$v= \frac{p}{m}= \frac{40120 kg m/s}{1180 kg}=34 m/s$

5 0

3 years ago

Read 2 more answers