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5 days ago

Dominic is the star discus thrower on South's varsity track and field team. In last year's regional competition, Dominic whirled

the 1.6 kg discus in a circle with a radius of 1.1m, ultimately reaching a speed of 52 m/s before launch. Determine the net force acting upon the discus in the moments before launch.

Physics

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In 2014, about how far in meters would you have to travel on the surface of the Earth from the North Magnetic Pole to the Geogra

Sav [3153]

Answer:

267.07 km

Explanation:

The given earth's radius is 6378.1 km

In the year 2014, the magnetic north pole was situated 2.40° away from the geographical north pole

2.40° $=2.40\times \frac{\pi }{180}=0.041866 radian$

The linear distance can be calculated using the formula $S=R\Theta =6378.1\times 0.041866=267.07km$

Thus, travelling from the magnetic north pole to the geographic north pole requires a distance of 267.07 km

3 0

1 month ago

Which table correctly identifies the abbreviation for SI units of length mass volume and temperature

Ostrovityanka [3204]

Response: b

Clarification:

Temperature: kelvin

Mass: kilogram

Length: meter

Volume: cubic meter

(Apex learning test)

8 0

1 month ago

A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximate

inna [3103]

Answer:

0.0031 m

Explanation:

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

$\lambda$ = Wavelength = 550 nm

d = Pupil diameter

$\theta$ = Angle

We have the expression

$tan\theta=\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{281\times 10^{-6}}{1.3}$

We have the expression

$sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow d=\dfrac{1.22\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1.22\times 550\times 10^{-9}}{sin\left(tan^{-1}\frac{281\times 10^{-6}}{1.3}\right)}\\\Rightarrow d=0.0031\ m$

The pupil diameter calculates to 0.0031 m

4 0

1 month ago

An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft

kicyunya [3294]

Response:

a) 80 V

b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( $K_{A}$ = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( $K_{B}$ = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

Solution

(a) Utilizing the given values for VA, $K_{B}$ and q, we derive a relationship among the three parameters and VB to compute VB.

At points A and B, the charge moves from A to B due to the electric field. The mechanical energy of the object remains conserved throughout this journey, allowing us to apply eq(1) in this context:

$K_{A} +U_{A} =K_{B} +U_{B}$ .........................................(1)

Where $K_{A}$ = 0, and the potential energy U of the charge is defined as U = q V

In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:

0+qVA= $K_{B}$ +qVB (Dividing by q)

VA= $K_{B}$ /q + VB (Restructuring for VB)

VB=VA- $K_{B}$ /q.......................................(2)

We now have the relation between VB, VA, and $K_{B}$ , allowing us to substitute our values for VA, $K_{B}$ , and q into equation (2) to obtain VB

VB=VA- $K_{B}$ /q

=30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)

=80 V

(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points

VA-VB = $\int\limits^1_0 {E} \, dl$ ...................................(a)

$=E\int\limits^1_0 {} \, dl$

VA-VB=E $l$ (Restructuring for E)

E= VA-VB/ $l$ ..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

E= VA-VB/ $l$

=-100 N/C

The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.

5 0

1 month ago