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bekas [8.4K]
3 years ago
6

Direction of wind: from to​

Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

uhm to what? i dunno what your talking about but someone else might

Explanation:

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A 4.9-MeV (kinetic energy) proton enters a 0.28-T field, in a plane perpendicular to the field. Part APart complete What is the
BartSMP [9]

Answer:

r=1.14m

Explanation:

\theta is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:

F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB

A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:

F_m=F_c\\qvB=F_c(1)

F_c is the centripetal force and is defined as:

F_c=m\frac{v^2}{r}

Here v is the proton's speed and r is the radius of the circular motion. Replacing this in (1) and solving for r:

qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}

Recall that 1 J is equal to 6.242*10^{12}MeV, so:

4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J

We can calculate v from the kinetic energy of the proton:

K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}

Finally, we calculate the radius of the proton path:

r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m

8 0
4 years ago
I NEED THE ANSWER ASAP I GIVE THANKS + 5 STARS
Kobotan [32]
Work or energy spent=force times distance=25N*4m=100Nm
power=energy or work divided by time=100Nm/5s=20Nm/s
5 0
4 years ago
A block of mass 0.404 0.404 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result, the spring
VARVARA [1.3K]

To solve this problem it is necessary to apply the concepts related to the Force from Hook's law as well as the definition of the period provided by the same definition.

We know that the Force can be defined as

F = xk \rightarrow mg = kx \Rightarrow k = \frac{mg}{x}

Where

k = Spring constant

x = Displacement

g = Gravity

m = mass

At the same time the period of a spring mass system is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Where

m = Mass

k = Spring constant

Our values are given as,

m = 0.404kg

x = 0.666m

Replacing to find the value of the Spring constant we have that

k = \frac{mg}{x}

k = \frac{(0.404)(9.8)}{0.666}

k = 5.944N/m

Now using the formula of the period we know that

T = 2\pi \sqrt{\frac{m}{k}}

T = 2\pi \sqrt{\frac{0.404}{5.944}}

T = 1.638s

Finally, if the oscillation was 0.359m

The maximum height will be determined by the total length of that oscillation being equivalent to

h=2a

h = 2*0.359

h = 0.718m

4 0
3 years ago
1. What are three examples of how invasive species spread?
Darya [45]

Answer:

Their primary way of spreading is from human activities, they can quickly travel around the world for example these new "murder" hornets that can kill a large bee hive with one sting, they traveled all the way from Asia. Invasive species can also be through people's luggage, small boats, planes and large shipment like cargo carriers. I hope this helps. :  )

Explanation:

8 0
3 years ago
What is the magnitude of a point charge in coulombs whose electric field 58 cm away has the magnitude 2.8 N/C?
Yuliya22 [10]

Answer:

q = 1.05 × 10^-10 C

Explanation:

E = 2.8 N/C

r = 58cm = 0.58m

k = 9×10^9

q = ?

E = f/q

E = kq/r²

q = E*r²/k

q = (2.8×0.58²)÷9×10^9

q = 1.05 × 10^-10 C

8 0
3 years ago
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