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kondaur [170]
3 years ago
6

The standard deviation of Eric’s data is 0.8°C. Martha conducted the same experiment. Her average temperature was 35.1 with a st

andard deviation of 1.2°C. Her data are more precise than Eric's
Physics
1 answer:
REY [17]3 years ago
5 0

Answer:

less precise than

Explanation:

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What does it mean that " memory is organized in semantic<br> networks
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7 0
2 years ago
A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft
elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0
3 years ago
To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal
tatiyna
Answer:  Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v =  linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
 Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).


5 0
2 years ago
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