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Leno4ka [110]
2 years ago
7

A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

squared in 10.8 seconds. What is his final velocity?
Physics
1 answer:
djyliett [7]2 years ago
5 0

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, u=32 m/s

Acceleration of the driver is, a=-1.5 m/s²

Time taken to reach final velocity is, t=10.8 s.

The final velocity is given using the Newton's equations of motion as:

v=u+at, where, v is the final velocity.

Now, plug in the given values and solve for v.

v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}

Therefore, his final velocity is 15.8 m/s.

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A fan connected to a 120-volt electrical system by an extension cord was worn through and exposed the bare, energized conductor,
kari74 [83]

Answer: the ladder

Explanation: since the energized conductor is already in contact with the ladder there by making electric current to flow. The base of the ladder is on the ground there by making the circuit to be complete and causing electrocution.

8 0
2 years ago
Read 2 more answers
A light horizontal spring has a spring constant of 138 N/m. A 3.35 kg block is pressed against one end of the spring, compressin
BARSIC [14]

Answer:

U_k = 0.113

Explanation:

using the law of the conservation of energy:

E_i -E_f=W_f

\frac{1}{2}Kx^2=NU_kd

where K is the spring constant, x is the spring compression, N is the normal force of the block, U_k is the coefficiet of kinetic friction and d is the distance.

Also, by laws of newton, N is calculated by:

N = mg

N = 3.35 kg * 9.81 m/s

N = 32.8635

So, Replacing values on the first equation, we get:

\frac{1}{2}(138)(0.123)^2= (32.8635)U_k(0.281m)

solving for U_k:

U_k = 0.113

8 0
3 years ago
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

<u>ac = 3.92 m/s²</u>

5 0
2 years ago
650 N boy and a 419 N girl sit on a 150 N porch swing that is 2.0 m long. If the swing is supported by a chain at each end, what
Sergio039 [100]

The tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

<h3>Principle of moments</h3>

The Principle of Moments states that when a body is in equip, the sum of clockwise moment about a point is equal to the sum of anticlockwise moment about the same point.

The formula for calculating moment is given below:

  • Moment = Force × perpendicular distance from the pivot

<h3>Calculating the tension in the chains</h3>

From the principle of moments:

Let tension in chain 1 be T1 and tension in chain 2 be T2.

T1 + T2 = 150 + 650 + 419

T1 + T2 =1219

Taking all distances from chain 1,

Sum of Moments = 0

419 × 0.5 + 150 × 0.85 + 650 × 0.9 = T2 × 1.7

T2 = 922/17

T2 = 542.35 N

Then, T1 = 1219 - 542.35

T1 = 676.65 N

Therefore, the tension in the two chains T1 and T2 is 676.65 N and 542.53 N respectively.

Learn more about tension and moments at: brainly.com/question/187404

brainly.com/question/14303536

7 0
2 years ago
A spherical weather balloon is filled with hydrogen until its radius is 4.40 m. Its total mass including the instruments it carr
ipn [44]

Answer:

4515.49484 N

4329.10484 N

Explanation:

r = Radius of balloon = 4.4 m

m = Mass of balloon with instruments = 19 kg

g = Acceleration due to gravity = 9.81 m/s²

Volume of balloon

v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 4.4^3\\\Rightarrow v=356.8179\ m^3

The Buoyant force = Weight of the air displaced

F=\rho vg\\\Rightarrow F=1.29\times 356.8179\times 9.81\\\Rightarrow F=4515.49484\ N

The buoyant force acting on the balloon is 4515.49484 N

Net force on the balloon

F_n=F-W\\\Rightarrow F_n=4515.49484-19\times 9.81\\\Rightarrow F_n=4329.10484\ N

The net force on the balloon is given by 4329.10484 N

As the balloon goes up the pressure outside reduces as the density of air decreases while the air pressure inside the balloon is high hence, the radius of the balloon tend to increase as it rises to higher altitude.

5 0
2 years ago
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