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Pavel [41]
3 years ago
7

A car accelerates from rest. It reaches a velocity of 25m/s in 10 seconds. What was the cars acceleration

Physics
1 answer:
soldier1979 [14.2K]3 years ago
7 0

Answer:

a = 2.5 m/s^2

Explanation:

u = 0

v = 25

t = 10

(using first eq. of motion)

a = (v - u) /t

a = (25 - 0) /10

a = 25/10

a = 2.5 m/s^2

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3 years ago
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Which example provides the most complete description of an object's motion?
cupoosta [38]

Answer:

The hiker followed a road heading north for 2 miles in 30 minutes.

Explanation:

In order to describe the motion of an object, distance covered and time taken must be required. The total path covered by an object is called the distance travelled.

The hiker followed a road heading north for 2 miles in 30 minutes. This describes the motion of hiker. The motion shows how fast the hiker is moving.  

Distance, d = 2 miles = 3218.6 m

times, t = 30 minutes = 1800 seconds

So, we can say that the hiker is moving with a speed of 1.78 m/s in north direction.

Hence, this is the required solution.

5 0
4 years ago
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Which method determines the relative age of Earth's sedimentary layers based on their organization?
MrRa [10]

Answer:

Relative age-dating involves comparing a rock layer or rock structure with other near-by layers or structures. Using the principles of superposition and cross-cutting relationships, and structures such as unconformities, one can determine the order of geological events.

7 0
3 years ago
A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski
Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

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from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

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6 0
3 years ago
g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the
Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

T^2 = \frac{4\pi^2 d^3}{2GM}

The given data are given with respect to known constants, for example the mass of the sun is

m_s = 1.989*10^{30}

The radius between the earth and the sun is given by

r = 149.6*10^9m

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

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m = 8.2*1.989*10^{30}

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Substituting in Kepler's third law:

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T = 120290789.7s

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T \approx 3.8143 years

Therefore the period of this star is 3.8years

7 0
3 years ago
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