What are the relative ages of the features in order of oldest to youngest?

A proton (mass m 1.67 10 27 kg) is being accelerated along a straight line at 3.6 1015 m/s2 in a machine. If the proton has an i

tankabanditka [31]

Answer:

(a) the speed is 2.93 × 10⁷ m/s

(b) the proton's kinetic increased by 2.36 × 10⁻¹³ J

Explanation:

The given information is:

initial speed, v_i = 2.4×10⁷ m/s
distance travelled, d = 0.035 m
acceleration, a = 3.6×10¹⁵ m/s²
mass, m = 1.67×10⁻²⁷ kg

(a) We must first determine the time it took the proton to travel the given distance:

t = d / v_i

t = (0.035 m) / (2.4×10⁷ m/s)

t = 1.46 × 10⁻⁹ s

Therefore, using the equation kinematics, we can determine the speed of the proton after 1.46 × 10⁻⁹ seconds. The speed is:

v = v_i + a t

v = (2.4 × 10⁷ m/s) + (3.6×10¹⁵ m/s²)(1.46 × 10⁻⁹ s)

v = 2.93 × 10⁷ m/s

(b) We must determine the inertial kinetic energy and the final kinetic energy:

The initial kinetic energy is:

EK_i = 1/2 m v_i²

= 1/2(1.67 × 10⁻²⁷ kg)(2.4 × 10⁷ m/s)²

= 4.81 × 10⁻¹³ J

The final kinetic energy is:

EK_f = 1/2 m v_f²

= 1/2(1.67 × 10⁻²⁷ kg)(2.93 × 10⁷ m/s)²

= 7.17 × 10⁻¹³ J

Therefore, the proton's kinetic increased by:

EK_f - EK_i = (7.17 × 10⁻¹³ J) - (4.81 × 10⁻¹³ J)

EK_f - EK_i = 2.36 × 10⁻¹³ J

7 0

3 years ago

Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

sweet [91]

Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

Electric field :

Electric field is proportional to the charge divided by square of distance.

In mathematically,

$E\propto\dfrac{q}{r^2}$

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

We need to find the electric field at black dot

According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

$E=\dfrac{kq}{r^2}$

The electric field will be largest.

(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{k2q}{(2r)^2}$

$E=\dfrac{kq}{2r^2}$

The electric field will be smallest.

(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.

Then, the electric field will be

$E=\dfrac{k2q}{(r)^2}$

The electric field will be very largest.

(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

Then, the electric field will be

$E=\dfrac{kq}{(2r)^2}$

$E=\dfrac{kq}{4r^2}$

The electric field will be very smallest.

So, The electric field from largest to smallest will be

$E_{3}>E_{1}>E_{2}>E_{4}$

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

4 0

3 years ago

The Sun appears at an angle of 48.4° above the horizontal as viewed by a dolphin swimming underwater. What angle does the sunlig

Luden [163]

Answer:

4.57 degree

Explanation:

As the dolphin is under water, so the angle is angle of refraction which is equal to 48.4 degree.

refractive index of water, n = 1.333

Let the angle of incidence which is the angle between the incident ray and the normal is i.

By use of Snell's law

$n = \frac{Sin i}{Sin r}$

$1.333 = \frac{Sin i}{Sin 48.4}$

$Sin i = 1.333\times Sin 48.4 = 0.9968$

i = 85.43 degree

Angle made with the horizon = 90 - 85.43 = 4.57 degree

6 0

3 years ago

The loudness of a sound is the wave's _______

Burka [1]

Amplitude:)

Hope this help!

5 0

3 years ago

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PLEASE HELP ME I NEED IT!!

Alla [95]

Answer:

d

Explanation:

7 0

3 years ago

Read 2 more answers