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givi [52]
2 years ago
10

A cyclist goes around a circular track once every 2 minutes. If the radius of the circular track is 105 meters calculate his spe

ed
Physics
2 answers:
Bad White [126]2 years ago
8 0
The formula should be (2πr)/T. r=radius T=time. if time is in seconds the formula would look like this
\frac{2\pi \times 105}{60 \times 2}
ankoles [38]2 years ago
5 0

Speed = (distance covered) / (time to cover the distance)

Distance covered = (π · diameter) = 210π meters

Time to cover the distance = 2 minutes or 120 seconds

Speed = (210π meters) / (120 seconds)

Speed = 5.5 meters/sec

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The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu
larisa [96]

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

5 0
2 years ago
Find your mass if a scale on earth reads 650 N when you stand on it.
netineya [11]

Weight = (mass) x (gravity)

Acceleration of gravity on Earth = 9.8 m/s²

                                           Weight on Earth = (mass) x (9.8 m/s²)

Divide each side by  (9.8 m/s²):          Mass = (weight) / (9.8 m/s²)

                                                            Mass = (650 N) / (9.8 m/s²)

                                                           Mass = 66.33 kg  (rounded)
 
7 0
2 years ago
Where do incident rays that are parallel to the principal axis converge to after reflecting off a concave mirror?
morpeh [17]

Answer:

At focus

Explanation:

A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.

These rays,  parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror

3 0
2 years ago
The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
snow_lady [41]
Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0
2 years ago
Read 2 more answers
If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.55 x 10-4 T) at a distance of 25
antiseptic1488 [7]

we are given in the problem the following dimensions or specifications 
B = 0.000055 T r = 0.25 m constant mu0 = 4*pi*10-7 

The formula that is applicable from physics is 
B = mu0*I/(2*pi*r) I = 2*B*pi*r/mu0 I = 68.75 Amperes 
7 0
2 years ago
Read 2 more answers
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