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anygoal [31]
4 years ago
8

A certain car going from Atlanta to Philadelphia averages 24.0 m/s over the course of the trip. Determine how long it takes the

car to reach Philadelphia if 1,250 km away from atlanta by road. Report your answer in hours. (Be careful of units!)
Physics
1 answer:
Dennis_Churaev [7]4 years ago
6 0

Answer:

14.5 hours

Explanation:

The relationship between speed, time and distance for a uniform motion is

v=\frac{d}{t}

where

v is the speed

d is the distance

t is the time taken

Here we know:

v = 24.0 m/s is the average speed

d=1250 km = 1.25\cdot 10^6 m is the distance

Solving the equation, we find the time taken:

t=\frac{d}{v}=\frac{1.25\cdot 10^6 m}{24.0 m/s}=52083 s

And since

1 hour = 3600 s

The time in hours is

t=52083 s \cdot \frac{1}{3600 s}=14.5 h

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A car’s tire rotates 5.25 times in 3 seconds. What is the tangential velocity of the tire?
Lena [83]

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire which is 11(r) m/s.

<h3>Angular velocity of the tire</h3>

The angular velocity of the tire is the rate of change of angular displacement of the tire with time.

The magnitude of the angular velocity of the tire is calculated as follows;

ω = 2πN

where;

  • N is the number of revolutions per second

ω = 2π x (5.25 / 3)

ω =  11 rad/s

<h3>Tangential velocity of the tire</h3>

The tangential velocity of the car's tire is the product of the angular velocity and radius of the car's tire.

The magnitude of the tangential velocity is caculated as follows;

v = ωr

where;

  • r is the radius of the car's tire

v = 11r m/s

Learn more about tangential velocity here: brainly.com/question/25780931

4 0
3 years ago
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m bef
bekas [8.4K]

Answer:

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

3 0
4 years ago
Help with 2 Physics questions, WILL CHOOSE BRAINLIEST
Tju [1.3M]

1) D

2) D.) Greater than \theta_c

Explanation:

1)

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

  • The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
  • The angle of incidence is greater than a certain angle called critical angle

In picture 1, we have 4 different diagrams. In the diagrams:

  • The red arrow represents the incident ray
  • The green arrow represents the refracted ray
  • The blue arrow represents the reflected ray

Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

2)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

D.) Greater than \theta_c

3 0
3 years ago
By applying a force of one Newton, one can hold a body of mass _​
SVEN [57.7K]

Answer:

By applying a force of one Newton, one can hold a body of mass of 102 gram.

Explanation:

  • Force is the pull or push of an object. It can be mathematically measured as, F= m* g.

                       where, F= force in newton

                                  m= mass in kg

                                  g= acceleration due to gravity (meter/second^{2} )

  • For 1 newton force,

                                   F= m* 9.8

                            or,   m= \frac{1}{9.8}= 0.102 kg

                            or,   m= 102 gram.

  • Hence, 102 gram mass can be hold by one Newton force.

3 0
4 years ago
Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

                     = 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

4 0
3 years ago
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