Speed / 16.0 m/s 12.0- 8.0 4.0 0 0 0.5 1.0 1.5 time / seconds 2.0 2.5 3.0 calculate the distance that the car travels in the first 0.5 seconds calculate the distance that the car travels in the first 0.5 seconds

Answers

Answer 1

Answer:

Answer: 8 m

Explanation:

From the equation distance = velocity * time, we can find the distance from a velocity vs. time graph by finding the area under the curve, since we get that area from multiplying velocity and time together.

For the first 0.5 seconds, the velocity is 16 m/s and the change in time is 0.5.

16 * 0.5 = 8 m.

Learn more about the relationship between distance and velocity here: brainly.com/question/29409777

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Radio waves travel at the speed of light, which is 3.00 u 108 m/s. How many kilometers will radio messages travel in exactly one year?

Answers

Answer:

Distance traveled will be $9.462* 10^(12)km$

Explanation:

We have given speed of the light $v=3* 10^8m/sec$

We have to find the distance traveled by light in 1 year

We know that 1 year = 8760 hour

And 1 hour = 60×60 = 3600 sec

So 1 year $=8760* 3600=3.154* 10^(7)sec$

We know that distance = speed × time

So distance traveled by light in one year $=3.154* 10^7* 3* 10^8=9.462* 10^(15)m$

We have to fond the distance in km

As we know that 1 km = 1000 m

So $9.462* 10^(15)m=(9.462* 10^(15))/(1000)=9.462* 10^(12)km$

Final answer:

Radio messages will travel approximately 9.46 x 10^12 kilometers in exactly one year.

Explanation:

Radio waves travel at the speed of light which is approximately 3.00 x 108 m/s.

To find out how many kilometers radio messages will travel in exactly one year, we need to first convert the speed of light into kilometers per second.

Since there are 1000 meters in a kilometer, we divide the speed of light by 1000 to convert from meters to kilometers. Then, we multiply this value by the number of seconds in a year (365 days x 24 hours x 60 minutes x 60 seconds) to get the total distance traveled in kilometers.

In summary, radio messages will travel approximately 9.46 x 1012 kilometers in exactly one year.

Learn more about Speed of light here:

brainly.com/question/32416355

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A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally away from and below the point of release. What is the speed of the ball just before it strikes the ground?

Answers

The speed of the ball just before it strikes the ground is equal to 46.55 m/s.

Given the following data:

Horizontal distance = 65 meters

Height of building = 0.10 km = 100 meters

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 $m/s^2$ .

To determine the speed of the ball just before it strikes the ground:

First of all, we would determine the time it took the ball to strike the ground by using the formula for maximum height.

$H = (1)/(2) gt^2\n\n100 = (1)/(2) * 9.8 * t^2\n\n200 = 9.8t^2\n\nt^2 = (200)/(9.8) \n\nt^2=20.41\n\nt=√(20.41)$

Time, t = 4.52 seconds

Next, we would find the horizontal velocity:

$Horizontal\;velocity = (horizontal\;distance)/(time) \n\nHorizontal\;velocity = (65)/(4.52)$

Horizontal velocity, V1 = 14.38 m/s

Also, we would find the velocity of the ball in the horizontal direction:

$V_2^2 = U^2 + 2aS\n\nV_2^2 = 0^2 + 2(9.8)(100)\n\nV_2^2 = 1960\n\nV_2 = √(1960) \n\nV_2 = 44.27 \;m/s$

Now, we would calculate the speed of the ball just before it strikes the ground by finding the resultant speed:

$V = √(V_1^2 + V_2^2) \n\nV = √(14.38^2 + 44.27^2)\n\nV = √(206.7844‬ + 1959.8329‬)\n\nV =√(2166.6173)$

Speed, V = 46.55 m/s

Answer:

v = 46.55 m/s

Explanation:

It is given that,

A ball is thrown horizontally from the top of a building 0.10 km high, d = 0.1 km = 100 m

The ball strikes the ground at a point 65 m horizontally away from and below the point of release, h = 65 m

At maximum height, velocity of the ball is 0. So, using the equation of motion as :

$d=ut+(1)/(2)at^2$

Here, a = g

$100=0+(1)/(2)* 9.8t^2$

$t=4.51\ s$

Let $v_x$ is the horizontal velocity of the ball. It is calculated as :

$v_x=(65\ m)/(4.51\ s)=14.41\ m/s$

Let $v_y$ is the final speed of the ball in y direction. It can be calculated as :

$v_y^2+u_y^2=2as$

$u_y=0$

$v_y^2=2gd$

$v_y^2=2* 9.8* 100$

$v_y=44.27\ m/s$

Let v is the speed of the ball just before it strikes the ground. It is given by :

$v=√(v_x^2+v_y^2)$

$v=√(14.41^2+44.27^2)$

v = 46.55 m/s

So, the speed of the ball just before it strikes the ground is 46.55 m/s. Hence, this is the required solution.

Which term best describes the change in frequency of waves when there is motion between the source of the waves and the observer?a. the Doppler effect
b. the Avogadro effect
c. the motor wave effect
d. the wave frequency effect

Answers

The Doppler effect is the term that best describes the change in frequency of waves when there is motion between the source of the waves and the observer. The correct option among all the options that are given in the question is the first option or option "a". I hope the answer helps you.

Answer:

Explanation:

F=ma make m the subject

Answers

Answer:

M=f/a

Explanation:

F=m x a

÷a both sides

F÷a = m

M=f/a

How many times more intense is an earthquake that measures 7.5 on the Richter scale than one that measures 6.7? (Recall that the Richter scale defines magnitude of an earthquake with the equation M=log i/s , where i is the intensity of the earthquake being measured, and S is the intensity of a standard earthquake

Answers

.5 - 6.7 = 0.8
10^0.8 ≅ 6.309573445
The 7.5 earthquake is approximately 6.3 times as intense as the 6.7 earthquake

How to figure it using your equation:
M = log(i/s)
10^M = i/s
i = s·10^M
ratio of the intensities of the two earthquakes = (s·10^7.5)/(s·10^6.7)
= 10^(7.5-6.7)
= 10^0.8
≅ 6.3:1

Answer:

6.31 times ( approx )

Explanation:

Given formula to find the magnitude of an earthquake is,

$M=log((I)/(S))$

Where,

I = intensity of the earthquakebeing measured,

S = theintensity of a standard earthquake,

By the above formula,

$10^(M)=(I)/(S)$

$\implies I = S* 10^(M)$

Since, S is constant,

If M = 7.5,

Then intensity of the earthquake,

$I_1=S* 10^(7.5)$

If M = 6.7,

Then intensity of the earthquake,

$I_2=S* 10^(6.7)$

$(I_1)/(I_2)=(S* 10^(7.5))/(S* 10^(6.7))=10^(7.5-6.7)=10^(0.8)\approx 6.31$

$\implies I_1=6.31I_2$

Hence, the earthquake that measures 7.5 on the Richter scale is 6.3 times more intense than the earthquake that measures 6.7.

You can find electric power lines under the ground by looking for magnetic fields at ground level. This is best explained by which of these statements?A) All wires have magnetic materials in them.
B) An electric current in a wire produces a magnetic field.
C) The power company puts magnets to help find the power lines.
D) The iron in the ground is disturbed by the power lines creating magnetic fields