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DENIUS [597]
2 years ago
14

A bicyclist is riding at a tangential speed of 13.2 m/s around a circular track. The magnitude of the centripetal force is 377 N

, and the combined mass of the bicycle and rider is 86.5 kg. What is the track's radius?​
Physics
1 answer:
MA_775_DIABLO [31]2 years ago
4 0

Answer:

40m approximately

Explanation:

Given

Force =377N

Mass =86.5kg

Velocity =13.2m/s

Required

Radius of the track

The expression for the centripetal force acting on the cyclist is

F=mv²/r

Make r subject of the formula

r= mv²/F

Substitute

r=86.5*13.2²/377

r= 15,071.76/377

r=39.97

r=40m approximately

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The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?
saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0
2 years ago
Not too long ago houses were protected from excessive currents by fuses rather than circuit breakers. sometimes a fuse blew out
olganol [36]

Answer: Resistance = 8.21 \times 10^{-8} \Omega

The approximate diameter of a penny is, <em>d</em> = 20 mm

thickness of penny is, <em>L = </em> 1.5×10^{-3} mm

The area of penny along circular face is,A = \frac{\pi d^2}{4} =\frac{\pi (20 mm\frac{1 m}{1000 m})^2}{4}

= 3.14×10^{-4} m²

The resistivity of copper is <em>ρ</em> = 1.72 x 10-8 Ωm.

Resistance,

R = \rho \frac{L}{A}  = (1.72 \times 10^{-8} \Omega m)\frac{1.5 \times 10^{-3} m}{3.14 \times 10^{-4} m^2} = 8.21 \times 10^{-8} \Omega

5 0
2 years ago
17. The edges of a rectangular solid have these measures: 1.5 feet by 1½ feet by 3 inches. What is its volume in cubic inches? a
Alisiya [41]

Answer:

c. 972

Explanation:

The volume of a rectangular solid is calculated as the product of its dimensions, that is, its width, its length and its height:

V=a*b*h

1 feet is equal to 12 inches, so:

a=1.5ft*\frac{12in}{1ft}=18in\\b=(1+\frac{1}{2})ft\\b=\frac{3}{2}ft*\frac{12in}{1ft}=18in

Now, we calculate the volume of the object in cubic inches:

V=18in*18in*3in\\V=972in^3

6 0
2 years ago
A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k
9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg  : mass of the  skier

θ =30°  :angle θ of the ski slope  with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W=  92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N=  0.10*780.8 N  

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 =  (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0
2 years ago
Seema knows the mass of basketball. What other information is needed to find the balls potential energy
Lelu [443]

Answer: The height (position) of the ball and the acceleration due gravity

Explanation:

In this case we are taking about gravitational potential energy, which is the energy a body or object possesses, due to its position in a gravitational field.  In this sense, this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.  

In the case of the Earth, in which the gravitational field is considered constant, the gravitational potential energy U will be:  

U=mgh  

Where:

m is the mass of the ball

g=9.8 m/s^{2} is the acceleration due gravity (assuming the ball is on the Earth surface)

h is the height (position) of the ball respect to a given point

Note the value of the gravitational potential energy is directly proportional to the height.

8 0
2 years ago
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