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Rasek [7]
2 years ago
12

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 m57.0 m . If the track is completely flat and the race car is traveling at a constant 31.5 m/s31.5 m/s (about 7.0×101 mph7.0×101 mph ) around the turn, what is the race car's centripetal (radial) acceleration?
Physics
1 answer:
crimeas [40]2 years ago
5 0

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(31.5\ m/s)^2}{57\ m}

a=17.4\ m/s^2

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

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A solid cylinder is radiating power. It has a length that is ten times its radius. It is cut into a number of smaller cylinders,
S_A_V [24]

Answer:

The total number of small cylinder = 7.

Explanation:

Lets take

Radius of the large cylinder = R

length = L

L = 10 R

The total area A = 2 π R² + π R L

The length of the small cylinder = l

The number of small cylinder = n

L = n l

The total area of small cylinders

A'=n (2 π R² + π R l)

As we know that emissive power given as

P = A ε σ T⁴

For large cylinder

P = A ε σ T⁴      -----------1

For small cylinders

P'=A' ε σ T⁴    ------2

From 1 and 2

Given that

P'= 2 P

A' ε σ T⁴ =2 A ε σ T⁴

A'=2 A       (All others are constant)

n (2 π R² + π R l) =(2 2 π R² + π R L)

n (2  R² +  R l) = (2  R² +  R L)

n(2R^2+R\times \dfrac{L}{n}) = 2(2R^2+RL)

L = 10 R

n(2R^2+R\times \dfrac{10R}{n}) =2 (2R^2+R\times 10R)

n(2+\dfrac{10}{n}) =2( 2+ 10)

2 n +10 = 2 x 12

2 n +10 = 24

2 n = 24 -10

2 n = 14

n = 7

The total number of small cylinder = 7.

3 0
2 years ago
A particle's position as a function of time t is given by r⃗ =(5.0t+6.0t2)mi^+(7.0−3.0t3)mj^.
nalin [4]
The given particle position is
\vec{r}(t)=(5.0t+6.0t^{2})\hat{i}+(7.0-3.0t^{3})\hat{j}

When t=5 s, the position vector is
\vec{r}_{1} = (5+150)\hat{i}+(7-375)\hat{j} = 175\hat{i}-368\hat{j}

The reference vector at t = 5 s is
\vec{r}_(2) = 0\hat{i}+7\hat{j}.

Let θ = the angle between the two vectors.
Then, by definition,
\vec{r}_{1}.\vec{r}_{2}=|r_{1}||r_{2}|cos\theta

|r₁} =  √[175²+(-368)²] = 407.4911
|r₂| = 7
\vec{r}_{1}.\vec{r}_{2}=(-368)*(7) = -2576

Therefore
θ = cos⁻¹ -2576/(407.4911*7) = cos⁻¹ -0.9031 = 154.57°

Answer: 154.57°
3 0
2 years ago
Read 2 more answers
What is the angular acceleration of the pencil when it makes an angle of 10.0 degrees with the vertical?
Alex787 [66]
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torque = I * alpha, where I is moment of inertia, alpha is angular acceleration. 

Thus, 0.098 / 0.000075 = 1306.666... rad / s^2 

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6 0
2 years ago
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A billiard ball with a mass of 1.5kg is moving at 25m/s and strikes a second ball with a mass of 2.3 kg that is motionless. Find
aliya0001 [1]

Answer:

The velocity of the second ball after the collision will be 16.3 m/s.

Step-by-Step Explanation:

Law of conservation of momentum states that the total momentum of an isolated remains the same before and after the collision.

Let: mass of the first ball = m1 = 1.5 kg

mass of the second ball = m2 =  2.3 kg

Momentum before the collision:

velocity of the first ball = v1 = 25 m/s

velocity of the second ball = v2 = 0 m/s

[tex] Initial momentum = m1v1 + m2v2 = 1.5*25 + 2.3*0 = 37.5 kgm/s [\tex]

Momentum after the collision:

velocity of the first ball = v1 = 0 m/s

velocity of the second ball = v2 = x

[tex]momentum after the collision= m1v1 + m2v2 = 1.5*0 + 2.3*x = 2.3*x[\tex]

According to law of conservation of momentum:

Initial momentum = Momentum after collision

37.5 = 2.3*x

⇒ x = 37.5/2.3

x = 16.3 m/s

4 0
2 years ago
Which example illustrates a chemical change?
ad-work [718]
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8 0
2 years ago
Read 2 more answers
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