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gavmur [86]
3 years ago
12

Noah Formula is in an airplane which is flying at constant speed in a circular course of radius 6500 m, circling O’Hare‘s airpla

ne prior to landing. Noah observed that the plane completes each round trip in 345 seconds (5 minutes & 45 seconds).
What is the speed of the airplane?
AND
What is the centripetal acceleration of the airplane?
Physics
1 answer:
algol [13]3 years ago
3 0

Answer: this one looks hard

Explanation:

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What is the mass of a car that weighs 19,000 N on earth?
Anna71 [15]
Divide by 9.8 I think
 so <span>1938.77551</span>
8 0
3 years ago
A solid wooden cube, 30cm on each side can be totally submerged in water if it is pushed downward with a force of 54N. What is t
Stels [109]

Answer:

the density of the wooden cube is 204.1 kg/m³

Explanation:

Given;

applied force, F = 54 N

length of each side of the solid wooden cube, L = 30 cm = 0.3 m

mass of the wooden cube is calculated as;

F = mg

m = F/g

m = 54/9.8

m = 5.51 kg

The volume of the wooden cube is calculated as;

V = L³

V = (0.3)³

V = 0.027 m³

The density of the wooden cube is calculated as;

ρ = m/V

ρ = (5.51 kg) / (0.027 m³)

ρ = 204.1 kg/m³

Therefore, the density of the wooden cube is 204.1 kg/m³

4 0
2 years ago
A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
3 years ago
If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?
PolarNik [594]

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

None of the choices says this.

6 0
3 years ago
What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s
Tanzania [10]
The spring scale will read 559 Newton's or 125.7 pounds.
4 0
3 years ago
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