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shtirl [24]
4 years ago
15

(Please help, will give thanks + brainliest answer)

Physics
1 answer:
Bond [772]4 years ago
8 0

Answer:

A) 1.9m/s

Explanation:

Use the equation v=d/t

The time is 10 seconds. The distance is more complicated because it’s a circle so u use the equation for circumference C=2•pi•r. So you just do the circumference over time and get the answer. Hope this helps

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The CERN particle accelerator is circular with a circumference of 7.0 km.
Contact [7]

Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

8 0
4 years ago
A hot air balloon is hovering in the air when it drops a 40 Kg food package to some lost golfers. If the package is dropped from
UNO [17]
We can calculate this with the law of conservation of energy. Here we have a food package with a mass m=40 kg, that is in the height h=500 m and all of it's energy is potential. When it is dropped, it's potential energy gets converted into kinetic energy. So we can say that its kinetic and potential energy are equal, because we are neglecting air resistance:

Ek=Ep, where Ek=(1/2)*m*v² and Ep=m*g*h, where m is the mass of the body, g=9.81 m/s² and h is the height of the body.

(1/2)*m*v²=m*g*h, masses cancel out and we get:

(1/2)*v²=g*h, and we multiply by 2 both sides of the equation

v²=2*g*h, and we take the square root to get v:

v=√(2*g*h)

v=99.04 m/s

So the package is moving with the speed of v= 99.04 m/s when it hits the ground. 
5 0
3 years ago
How much total surface of the moon is illuminated by the sun when it is at quarter phase?
alex41 [277]
50% of the moon is always illuminated, however during it's quarter phase means that we only see a quarter of what's really lit up. So it LOOKS like the moon is only 25% lit and 75% dark, it's truly 50/50. We only see that 25% since we can see it from one angle.
6 0
3 years ago
A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is
Alika [10]

Answer: 24.97 kg

Explanation:

The gravitational force between two objects of masses M1, and M2 respectively, and separated by a distance R, is:

F = G*(M1*M2)/R^2

Where G is the gravitational constant:

G = 6.67*10^-11 m^3/(kg*s^2)

In this case, we know that

R = 0.002m

F = 0.0104 N

and that M1 = M2 = M

And we want to find the value of M, then we can replace those values in the equation to get

0.0104 N = (6.67*10^-11 m^3/(kg*s^2))*(M*M)/(0.002m)^2

(0.0104 N)*(0.002m)^2/(6.67*10^-11 m^3/(kg*s^2)) = M^2

623.69 kg^2 = M^2

√(623.69 kg^2) = M = 24.97 kg

This means that the mass of each object is 24.97 kg

6 0
3 years ago
A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by
lana66690 [7]

Answer:

Angle: 48.19^o

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

v_b\ cos\alpha

where v_b is the speed of the boy in still water and \alpha is the angle respect to the shoreline. If the river flows at speed v_s, we now set

v_b\ cos\alpha=v_s

\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}

\alpha=48.19^o

8 0
3 years ago
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