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Alik [6]
2 years ago
15

A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f

riction force of 250 N perpendicular to the orbit radius. What is the torque on the star?
Physics
1 answer:
snow_tiger [21]2 years ago
4 0

Answer:

2.5 \times 10^{11} N-m upwards        

Explanation:

Torque is the vector cross product of the force and radial distance.

\tau = rF sin \theta

\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

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A shot-putter exerts a force of 0.142 kN on a shot, accelerating it to 22.75 m/s2. What is the mass of the shot?
Svetach [21]

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Force and Power.

Since, according to the Newton's law,

Force = mass * Acceleration.

hence, here

Force = 142 N, accelration = 22.75 m/s2

hence, mass = 142/22.75

===> Mass = 6.24 Kg

hence the mass of the shot is 6.24 Kg

8 0
2 years ago
If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the induc
mel-nik [20]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

5 0
2 years ago
The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=4900
Sedaia [141]

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}.

(b) Now horizontally,

L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.

3 0
2 years ago
How much would it cost to cover the entire land area of United States with dollar bills? To answer this question, you may find t
s2008m [1.1K]

Answer:

It would cost approximately $925,455,484 million to cover continental US and Alaska with $1 bills.

Explanation:

the area of a one dollar bill = 6.5 cm x 15.5 cm = 100.75 sq cm

the approximate area of continental US + Alaska = (1,000 miles x 3,000 miles) x 1.2 = 3,600,000 sq miles

each sq mile is roughly 2.58999 sq km, so the total area in sq km = 9,323,964 sq km

1 sq km = 1,000,000 sq meters

each sq meter = 10,000 sq cm

1 sq m = 10,000 / 100.75 = 99.25558 bills

1 sq km = 99,255,583.13 bills

9,323,964 sq km = 925,455,483,900,000 bills

8 0
2 years ago
Read 2 more answers
Which statements describe intensity? Check all that apply.
REY [17]

Answer:

it's A, D, E

Explanation:

6 0
2 years ago
Read 2 more answers
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