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horsena [70]
3 years ago
9

Solve this worksheet. Need this worksheet with Answer

Physics
2 answers:
Ray Of Light [21]3 years ago
8 0
There’s nothing, it’s blank
andrey2020 [161]3 years ago
3 0
There’s nothing there? it’s all blank..
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Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

Plug values in

88.4^2+174^2=c^2

Simplify

7814.56+30276=c^2

Add the two values

38090.56=c^2

Take the square root of both sides

195.168\approx195.168

8 0
2 years ago
A feris wheel is turning at a constant speed of 5m/s is it not accelerating ,true or false
erma4kov [3.2K]

Answer:no it is staying the same speed

Explanation:

6 0
3 years ago
An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

6 0
3 years ago
Please help ASAP will mark brainliest How do we find the amount of uncertainty we should expect in our measurement of the period
kicyunya [14]

Answer:

The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.

Explanation:

hope it helps :)

4 0
3 years ago
What is the product of nuclear fission?
RideAnS [48]
Nuclear fission is seperating an atom so product is b.

Energy is released during nuclear fussion. Mass is converted into energy because E=mc^2, this is the energy that is released
5 0
3 years ago
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