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nikdorinn [45]
3 years ago
5

A car heading north collides at an intersection with a truck of the same mass as the car heading east. If they lock together and

travel at 28 m/s at 37° north of east just after the collision, how fast was the car initially traveling? Assume that any other unbalanced forces are negligible.
what is the correct answer
34 m/s
17 m/s
26 m/s
68 m/s
...?
Physics
1 answer:
stealth61 [152]3 years ago
3 0

This phenomenon is an inelastic collision. The problem is easier to solve because the masses of the colliding bodies are equal. By conservation of momentum, the final momentum of the bodies as one is equal to the VECTOR sum of the momentum of individual bodies. 

For this problem, the final velocity of the two cars is the sum of the initial velocities of each car. The velocity of the car is just the component of the velocity along the north direction.

Using trigonometry,
v1 = vf * sin(theta)
v1 = 28 * sin(37)

v1 = 16.9 m/s
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When an exothermic reaction releases thermal energy, this energy is usually...

C. Harnessed to do work
(Not 100% on that)
7 0
3 years ago
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2. If 66.2 J of thermal energy is removed from a 0.141 kg piece of iron, what is its change in temperature? The specific heat ca
yawa3891 [41]

Answer: I think is -5

Explanation:

3 0
3 years ago
Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v
mihalych1998 [28]

Answer:

a.

\displaystyle a(0 )=8.133\ m/s^2

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=0.52\ m/s^2

b.\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. t=9.9 \ sec

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

\displaystyle V(t)=a(1-e^{bt})

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

\displaystyle a=11.81\ ,\ b=-0.6887

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

\displaystyle V(t)=11.81(1-e^{-0.6887t})

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})

\displaystyle a(t)=8.133547\ e^{-0.6887t}

For t=0

\displaystyle a(0)=8.133547\ e^o

\displaystyle a(0 )=8.133\ m/s^2

For t=2

\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}

\displaystyle a(2)=2.05\ m/s^2

\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}

\displaystyle a(4)=0.52\ m/s^2

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C

\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

\displaystyle x(0)=0=>11.81\times1.45201+C=0

Solving for C

\displaystyle c=-17.1482\approx -17.15

Now we complete the equation for the distance

\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100

Rearranging

\displaystyle t+1.45\ e^{-0.6887t}=9.92

We define an auxiliary function f(t) to help us find the value of t.

\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92

Let's try for t=9 sec

\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92

Now with t=9.9 sec

\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184

That was a real close guess. One more to be sure for t=10 sec

\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).  

At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}

7 0
3 years ago
Consider the following three statements: (i) For any electro-magnetic radiation, the product of the wavelength and the frequency
Scilla [17]

Answer:

A and B

Explanation:

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength

c is the speed of light having value 3\times 10^8\ m/s

Thus, the product of the wavelength and the frequency is constant and equal to 3\times 10^8\ m/s

<u>Option A is correct.</u>

Given, Frequency = 1\times 10^{18}\ Hz

Thus, Wavelength is:

Wavelength=\frac{c}{Frequency}

Wavelength=\frac{3\times 10^8}{1\times 10^{18}}\ m

Wavelength=3\times 10^{-10}\ m

Also, 1 m = 3\times 10^{-10} Å

So,

<u>Wavelength = 3.0 Å</u>

<u>Option B is correct.</u>

As stated above, the speed of electromagnetic radiation is constant. Hence, each radiation of the spectrum travels with same speed.

<u>Option C is incorrect.</u>

3 0
3 years ago
Read 2 more answers
How long does it take anya to cover the distance of 5.00 miles ?
sammy [17]
Here is the full information about the question. <span>Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:
</span><span>
t = d / s </span>
<span>t = 2.5 (half of the total distance) / 8.5 (speed of running) </span>
<span>This is .294 hours which is about 1058s... </span>

<span>for the walking part... </span>

<span>t = d / s </span>
<span>t = 2.5 / 3.5 </span>
<span>t = 5/7hours = 2571 s. </span>
5 0
3 years ago
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