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2 years ago

A police officer draws a sketch of the scene of an accident, as shown.

Physics

2 answers:

iren2701 [21]2 years ago

5 0

I would have to say that it is Y

larisa86 [58]2 years ago

4 0

Answer:

Arrow Y

Explanation:

It is given that, the car was rounding the curve when it hit an oil slick and went off the road.

It is clear that the car is moving in a circular track. The force acting on a circular path is centripetal force. Also, the speed remains constant and the velocity changes in the case of circular motion. The direction of motion when the car leaves circular motion will be tangential.

Arrow Y shows the direction that the car moved in after hitting the oil slick.

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Three wires are made of copper having circular cross sections. Wire 1 has a length l and radius r. Wire 2 has a length l and rad

Alex73 [517]

Explanation:

Below is an attachment containing the solution.

4 0

2 years ago

Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o

ipn [44]

Answer:

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence, E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

U = E_p + E_k + E_g

U = 0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0

2 years ago

Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter

Mila [183]

The frequency of the radio wave is:
$f=102.1 MHz = 102.1 \cdot 10^6 Hz$

The wavelength of an electromagnetic wave is related to its frequency by the relationship
$\lambda= \frac{c}{f}$
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
$\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m$
and this is the wavelength of the radio waves in the problem.

7 0

2 years ago

A jogger accelerates from rest to 3.0 m/s in 2.0 s. A car accelerates from 38.0 to 41.0 m/s also in 2.0 s. (a) Find the accelera

Tom [10]

Answer:

a) The acceleration of the jogger is 1.5 m/s²

b) the acceleration of the car is also 1.5 m/s²

c) Yes, the car travels 76 m farther than the jogger.

Explanation:

a) The acceleration of an object is the variation of its velocity over time:

a = final velocity - initial velocity / time

for the jogger:

a = 3.0 m/s - 0 m/s / 2.0 s = 1.5 m/s ²

b) For the car:

a = 41.0 m/s - 38.0 m/s / 2.0 s = 1.5 m/s²

c) Let´s see the position of the car after 2 seconds.

The equation for the position of an accelerated object moving in a straight line is:

x = x0 + v0* t +1/2 * a * t²

Where:

x = position of the car at time "t"

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Let´s consider x0 = 0 because the origin of the reference system is located where the car starts accelerating. Then:

x = 38,0 m/s * 2 s + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 79 m

In the same way, we can calculate the position of the jogger:

x = 0 m/s * t + 1/2 * 1.5 m/s ² * (2.0 s)²

x = 3 m

The car travels 79 m - 3 m = 76 m farther than the jogger

4 0

2 years ago

A 1100kg car pulls a boat on a trailer. (a) what total force resists the motion of the car, boat,and trailer, if the car exerts

likoan [24]

Refer to the figure shown below.

m₁ = 1100 kg, the mass of the car
m₂ = 700 kg, the mass of the trailer and boat
F = 1900 N, the driving force acting on the car
N₁ = m₁g, the normal reaction on the car
N₂ = m₂g, the normal force on the trailer and boat
μN₁ and μN₂ are frictional forces, where = kinetic coefficient of friction
T = the force in the hitch between the car and trailer.

Part (a)
Let R₁ = the total force that resists the motion of the car, boat, and trailer.
Because the acceleration is 0.550m/s², therefore
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100+700 kg)*(0.55 m/s²) = (1900 - R₁) N
990 = 1900 - R
R₁ = 910 N

Answer: The resistive force is 910 N

Part (b)
80% of the resistive forces are experienced by the boat and trailer.
Let the resistive force be R₂.
Then
R₂ = 0.8*R₁ = 728 N
If the tension in the hitch between the car and the trailer is T, then
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²)
T - 728 = 385
T = 1113 N

Answer: The force in the hitch is 1113 N

3 0

2 years ago