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mote1985 [20]
4 years ago
10

What of the following are not examples of data

Physics
2 answers:
Masja [62]4 years ago
6 0
Do the work the fam there are in shortcuts in life
Serga [27]4 years ago
5 0
Theories are not examples of data
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a supertanker traveling at 7.2 m/s decelerates to a halt in 12 min. Calculate the magnitud of its average decelaration in meters
Nataliya [291]

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (0 m/s − 7.2 m/s) / (12 min ×  60 s/min)

a = -0.01 m/s²

8 0
3 years ago
A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. the flywheel has mass 40.0 kg and
tangare [24]
<span>The flywheel is solid cylindrical disc. Moment of inertial = ½ * mass * radius^2 Mass = 40.0 kg Radius = ½ * 76.0 cm = 38 cm = 0.38 meter Moment of inertial = ½ * 41 * 0.36^2 Convert rpm to radians/second The distance of 1 revolution = 1 circumference = 2 * π * r The number of radians/s in 1 revolution = 2 * π 1 minute = 60 seconds 1 revolution per minute = 2 * π radians / 60 seconds = π/30 rad/s Initial angular velocity = 500 * π/30 = 16.667 * π rad/s 170 revolutions = 170 * 2 * π = 340 * π radians The flywheel’s initial angular velocity = 16.667 * π rad/s. It decelerated at the rate of 1.071 rad/s^2 for 48.89 seconds. θ = ωi * t + ½ * α * t^2 θ = 16.667 * π * 48.89 + ½ * -1.071 * 48.89^2 2559.9 - 1280 θ = 1280 radians</span>
4 0
3 years ago
How much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 12 ohms?
Reil [10]

V = IR

Where v is voltage I is current and r is resistance

So

V = 9

R = 12

V/R = I

9/12 = I

I = 0.75 A

4 0
3 years ago
Read 2 more answers
At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the sta
aleksley [76]

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

<u>Given  :</u>

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

 =-4.8*10^22 N*p

<em>As this force is in one direction, we could get its vector as next  </em>

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.

8 0
3 years ago
A car collides head on with a stationary sign. The collision brings the car to a stop, but the sign does not move.
olga_2 [115]

A.a sign that breaks loose from the ground when a force is applied to it

7 0
3 years ago
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