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Novosadov [1.4K]
3 years ago
6

A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 15.0 min at 50.0 km/

h, 19.0 min at 85.0 km/h, and 55.0 min at 60.0 km/h and spends 25.0 min eating lunch and buying gas.
a. What is the total distance traveled over the entire trip (in km)?
b. What is the average speed for the entire trip (in km/h)?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer

given

Case 1-Speed of car 50 Km/h for 15 min

Case 2-Speed of car 85 Km/h for 19 min

Case 3 - Speed of car 60 km/h for 55 min

lunch break time = 25 min

a) total distance travel

   we know,

    distance = speed x time

 D = 50 \times \dfrac{15}{60} + 85 \times \dfrac{19}{60} + 60\times \dfrac{55}{60}

  D = 94.42 Km

b) average\ speed = \dfrac{total\ distance}{total\ time}

   average\ speed = \dfrac{94.42}{\dfrac{15}{60}+\dfrac{19}{60}+\dfrac{55}{60}+\dfrac{25}{60}}

average\ speed = 49.69\ km/h

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Answer:

So the sound intensity level they would experience without the earplugs is 110.32dB.

Explanation:

Given data

Sound intensity by factor =215

Sound intensity level =87 dB

To find

Sound intensity level they would experience without the earplugs

Solution

First we need to find the new sound intensity level

So

I_{n}=215(10^{\frac{87}{10} } )\\I_{n}=1.08*10^{11})

The dB can be calculated as:

dB=10log(I_{n})\\

Substitute the given values

dB=10log(1.08*10^{11})\\dB=110.32dB

So the sound intensity level they would experience without the earplugs is 110.32dB.

7 0
3 years ago
Today's topic
babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

A body is said to be in uniform motion if it covers equal distances in equal interval of time.

A body is said to be in non-uniform motion if it covers unequal distances in equal interval of time or vice-versa

4 0
2 years ago
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in
wel

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

3 0
3 years ago
Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f
mylen [45]

Answer:

The force will be zero

Explanation:

Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.

7 0
3 years ago
Find the value of F1 + F2 + F3.<br>​
Dovator [93]

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

<u>For F₃</u>

<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]

Now using the Pythagorean theorem we can find the total force.

F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]

8 0
3 years ago
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