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charle [14.2K]
3 years ago
10

an object starts at rest and accelerates until it is moving at 247 km/h. if it takes 1.7minutes for this change to occur what is

the average acceleration in m/s
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0

The average acceleration in 0.67 m/s^2.

The average acceleration is calculated as change in velocity/time

a=(247*5/18)/(1.7*60)

=68.61/102=0.67 m/s^2

Therefor the average acceleration of the object is 0.67 m/s^2

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EleoNora [17]

Explanation:

A. White Dwarf: It is the end stage of a Sun-like star. At the end the red giant star will throw away the outer layers in the form a shell and in the center a hot and bright core will remain. This core consists of carbon and oxygen but the conditions are not suitable to start the fusion of carbon and oxygen.

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5 0
3 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
A kid drives 4 miles to the mall. If the speed limit is 45 miles/hr and the kid makes the trip in .03 hours. Is the kid breaking
yuradex [85]

Answer:B

Explanation:

.03 of an hr is 2 mins and if it takes 2 mins to drive 4 miles enter it in pace calulator you are going 120mph so the closest to 120 is 133mph

7 0
3 years ago
11. A fundamental property of light is that it: 15
myrzilka [38]

Answer:

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Explanation:

Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.

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3 years ago
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vovangra [49]

YEP ITS D 2 N in Daniel's direction

8 0
3 years ago
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