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Sergio039 [100]
3 years ago
6

Bx = -1.33 m and By = 2.81 m Find the magnitude of the vector.

Physics
1 answer:
NISA [10]3 years ago
3 0

Answer:

Explanation:

The formula for the magnitude of a vector is

B_{mag}=\sqrt{(-1.33)^2+(2.81)^2} and then round to the hundredths place:

3.11 m. Since we are in Q2, we can also find the direction of this vector:

tan^{-1}(\frac{2.81}{-1.33})=-64.7 but since we are in Q2, we add 180 degrees to the result, getting the angle to be 115.3

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A positively charged particle initially at rest on the ground accelerated upward to 100m/s in 2.00s. If the particle has a charg
VashaNatasha [74]

Answer:

Explanation:

From the question we are told that

     The initial velocity is  u  =  100 m/s

       The time taken is  t = 2.0 s

       The charge to mass ratio is  Q/m  =  0.100 C/kg

       

Generally the acceleration is mathematically evaluated as

              a  =   \frac{u}{t }

substituting values  

               a =   \frac{100}{2}

                a =  50 \ m/s^2

The electric field is mathematical represented as

             E = \frac{(a+g)}{Q/m}

substituting values

              E = \frac{(50+9.8)}{0.100}

              E = 598 \ N/C

4 0
3 years ago
Does changing amplitude, wavelength or frequency play a role in how bluetooth acts?
Mandarinka [93]

Answer:

Yes

Explanation:

Yes, bluetooth devices work in a frequency range between 2.4 - 2.485GHz. Outside this frequency the devices will not communicate with each other correctly. This frequency equals a wavelength of around 1cm. Therefore, any change in the amplitude or wavelength would need to be in relation to each other in order to maintain the frequency in the required range for the bluetooth device to work accordingly. If one increases while the other remains the same it can easily change the frequency to outside the range.

3 0
3 years ago
Hola tengo un taller de física y nose como resolver este pregunta
Blizzard [7]

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

8 0
3 years ago
How many atoms are in a lightbulb
Gre4nikov [31]
It depends on the size, material, shape, etc.
8 0
3 years ago
A particular automotive wheel has an angular moment of inertia of 12 kg*m^2, and is decelerated from 135 rpm to 0 rpm in 8 secon
lozanna [386]

Answer:

(A) Torque required is 21.205 N-m

(b) Wok done will be equal to 1199.1286 j

Explanation:

We have given moment of inertia I=12kgm^2

Wheel deaccelerate from 135 rpm to 0 rpm

135 rpm = 135\times \frac{2\pi }{60}=14.1371rad/sec

Time t = 8 sec

So angular speed \omega _i=135rpm and \omega _f=0rpm

Angular acceleration is given by \alpha =\frac{\omega _f-\omega _i}{t}=\frac{0-14.1371}{8}=--1.7671rad/sec^2

Torque is given by torque \tau =I\alpha

=12\times 1.7671=21.205N-m

Work done to accelerate the vehicle is

\Delta w=K_I-K_F

\Delta W=\frac{1}{2}\times 12\times 14.137^2-\frac{1}{2}\times 12\times0^2=1199.1286J

8 0
4 years ago
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