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lora16 [44]
3 years ago
7

A electron is a subatomic particle that carries a ___________ charge

Physics
1 answer:
Aloiza [94]3 years ago
6 0
Negative charge hope this helped you
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4. Jimmy dropped a 10 kg bowling ball from a building that is 25 meters high.
KIM [24]

Answer:

The K.E of the bowling ball right before it hits the ground, K.E = 2450 J            

Explanation:

Given data,

The mass of the bowling ball, m = 10 kg

The height of the building, h = 25 m

The total mechanical energy of the body is given by,

                                     E = P.E + K.E

At height 'h' the P.E is maximum and the K.E is zero,

According to the law of conservation of energy, the K.E at the ground before hitting the ground is equal to the P.E at 'h'

Therefore, P.E at 'h'

                                  P.E = mgh

                                         = 10 x 9.8 x 25

                                         =  2450 J

Hence, the K.E of the bowling ball right before it hits the ground, K.E = 2450 J                                                                      

3 0
3 years ago
Which objects are inclined planes? Check all that apply.
Gre4nikov [31]
Wheel chair ramp, slide, and staircase
8 0
4 years ago
Read 2 more answers
Will binoculars work properly if their prisms (assume n = 1.50) are immersed in water (nwater= 1.33)? assume that binoculars wor
Amanda [17]

The binoculars will not work properly.

Please refer to the diagram attached.

Binoculars work on the principle of total internal reflection. The incident ray from air enters the prism through on face normally and it strikes the opposite face at an angle of 45°. The ray undergoes Total internal reflection if its critical angle is less than the angle of incidence,which , in this case is 45°.

The critical anglei_{c} is related to the refractive index of the material of glass μ as shown below.

\mu =\frac{1}{sini_c}

For a refractive index equal to 1.5, the critical angle is found as shown below.

sini_c=\frac{1}{\mu} \\ =\frac{1}{1.5} \\ =0.666

Take the inverse of the value 0.666 to determine the critical angle.

sin^{-1} (0.666)=41.8^o

The critical angle in air is less than the angle of incidence, and hence total internal reflection occurs in air.

When the binoculars are immersed in water, the ray passes into the glass through water. therefore, the refractive index of the prism when immersed in water is given by,

\mu_g_w=\frac{\mu_g}{\mu_w}

Therefore the critical angle <em>c</em> in this case is given by,

sinc=\frac{\mu_w}{\mu_g}

Substitute 1.33 for \mu_w and 1.5 for \mu_g.

sinc=\frac{\mu_w}{\mu_g} \\ =\frac{1.33}{1.5} \\ =0.8866

Take the inverse of the value 0.8866.

c=sin^{-1} (0.8866)\\ =62^o

Since the critical angle of the prism when immersed in water, is 62°, which is greater than the angle of incidence of 45° required for viewing the object, the binoculars which are set for Total reflection in air, will not function when immersed in water.

4 0
4 years ago
What is not a raquet sport
uysha [10]

Answer:

what

Explanation:

Racket sports include tennis, badminton, squash or any other sport where you use rackets to hit a ball or shuttlecock to play. They can be played competitively or just for fun and are a great form of physical activity.

7 0
3 years ago
(a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less
Free_Kalibri [48]

Answer:

q = 7.4 10⁻¹⁰ C

Explanation:

a) The magnetic force is given by the expression

        F = q v x B

Where the blacks indicate vectors, q is the electric charge, v at particle velocity and B the magnitude of the magnetic field. If the velocity is perpendicular to the magnetic field, the sine is 1

      F = q v B

Let's calculate the charge

      q = F / vB

      q = 1.00 10⁻¹² / 30.0 B

For the magnetic field of the earth we have a value between 25μT and 65μT, an intermediate value would be 45 μT, let's use this value.

     q = 1 10⁻¹² / (30 45 10⁻⁶)

    q = 7.4 10⁻¹⁰ C

b) In laboratories and modern electronics, currents of up to 1 10⁻⁶ A can be achieved without much difficulty, in advanced and research laboratories currents of up to 1 10⁻¹² can be managed. Load values ​​(coulomb) cannot they are widely used today for work, but 1 mA = 3.6C, so we see that getting loads with the value of 10⁻¹⁰ C implies very small current less than 1 10⁻¹³ A, which only in laboratories of Very specialized can be created. Consequently, from the above it would be difficult to find loads lower than the calculated

The electrostatic charge is the one created by the friction between two surfaces, it is an indicated charge, in this case it would be possible to have better wing loads found from 10⁻¹⁰C

4 0
3 years ago
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