Is a cremated flower pot a good conductor? please help.

the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a

inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

The initial velocity of the car, u = 0 m/s

The displacement of the ca, S = 1100 m

The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

S = ut + ½ at²

Substituting the given values,

1100 = 0 + ½ x a x 14²

a = 11.22 m/s²

Using the III equation of motion

v² = u² + 2 as

v = √(2as) (∵ u = 0)

Substituting,

v = √(2 x 11.22 x 1100)

= 157.11 m/s

The average speed of the car,

$V=\frac{0+65+157.11}{3}$

V = 74.04 m/s

Hence, the average velocity of the car is, V = 74.04 m/s

4 0

3 years ago

What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

azamat

Explanation:

Q1) What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?

Ans1) speed, v=st=2πrT=2×227×7×10-260×60=119×10-4=1.22×10-4m/s

<h2><em><u>Hope it helps</u></em></h2>

5 0

3 years ago

A 5 N force pushes on the right side of a box. At the same time, a 10 N force pushes on the left side of the box. What happens t

notka56 [123]

The resultant force is 5N.So the box moves to right with constant acceleration.

8 0

4 years ago

Read 2 more answers

A proton and an electron are placed in an electric field. Which undergoes the greater acceleration?

iren2701 [21]

Newton's 2nd law:

Fnet = ma

Fnet is the net force acting on an object, m is the object's mass, and a is the acceleration.

The electric force on a charged object is given by

Fe = Eq

Fe is the electric force, E is the electric field at the point where the object is, and q is the object's charge.

We can assume, if the only force acting on the proton and electron is the electric force due to the electric field, that for both particles, Fnet = Fe

Fe = Eq

Eq = ma

a = Eq/m

We will also assume that the electric field acting on the proton and electron are the same. The proton and electron also have the same magnitude of charge (1.6×10⁻¹⁹C). What makes the difference in their acceleration is their masses. A quick Google search will provide the following values:

mass of proton = 1.67×10⁻²⁷kg

mass of electron = 9.11×10⁻³¹kg

The acceleration of an object is inversely proportional to its mass, so the electron will experience a greater acceleration than the proton.

6 0

4 years ago

A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0

3 years ago