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lina2011 [118]
2 years ago
15

Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

ately 57.0 m . If the track is completely flat and the race car is traveling at a constant 30.5 m/s (about 68 mph ) around the turn, what is the race car's centripetal (radial) acceleration
Physics
1 answer:
umka21 [38]2 years ago
7 0

Answer:

a=16.32\ m/s^2

Explanation:

Given that,

The radius of the track, r = 57 m

The speed of a race car, v = 30.5 m/s

We need to find the centripetal acceleration of the car. Its formula that is use to find it is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(30.5)^2}{57}\\\\=16.32\ m/s^2

So, the car's centripetal acceleration is 16.32\ m/s^2.

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40 meters divided by 5
denis23 [38]

Answer:

8 meters

Explanation:

40/ 5 = 8

8 0
2 years ago
A car accelerates from 30 m/s to 50 m/s in 2 seconds. Calculate the cars acceleration
masha68 [24]

Answer:

applying 1st eq of motion vf=vi+at we have to find a=vf-vi/t here a=50-30/2=10 so we got a=10m/s²

5 0
3 years ago
Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it
olga2289 [7]
<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>
6 0
3 years ago
Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of
max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

Q =\frac{KA(\delta T)}{L}

Q =\frac{25.87*1*20}{1}

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0
3 years ago
Read 2 more answers
On the way to school, Jed traveled 100m north, 300m east, 100 north, 100m east, and 100m north. A.) Find the total distance trav
Oduvanchick [21]

Answer:

Total distance = 700 m

Displacement = 500 m

Explanation:

Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is:  300 + 400 = 700 m.

But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:

displacement = \sqrt{300^2+400^2} =\sqrt{250000} =500\,m

5 0
3 years ago
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