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iVinArrow [24]
3 years ago
9

An object began moving at the starting point (0,0) on the graph. The average speed of the object for the entire trip of 5.0 hour

s is:
A. 15mph
B. 17mph
C. 19mph
D. 20.0mph
E. 22mph

Physics
2 answers:
AURORKA [14]3 years ago
8 0
B 17


85 ÷ 5 would be the equation you do to solve
inna [77]3 years ago
4 0

Answer: So in your graph, you can see that after 5 hours, the object is in 85m. So your displacement over the lapse of 5 hours in 85m (because the object starts at 0m), the average speed will be distance over time, so 85m/5h=17m/h

So the correct answer is B, 17 meters per hour.

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Below is an "oracle" function. An oracle function is a function presented interactively. When you type in a t value, and press t
Bingel [31]

Answer:

The rate of change of height with respect to time is -10.64 feet/sec

Explanation:

Given that,

There are three lines, so you can calculate three different values of the function at one time.

The function f(t) represents the height in feet of a ball thrown into the air, t seconds after it has been thrown.

Given table is,

Time t = 0, 1, 1,02

Function is,F(t)=-3.053113177191196\times10^{-18},6.000000000000134, 6.41760000000015

We need to calculate the initial height of ball

Using equation of motion

h=h_{0}+ut-\dfrac{1}{2}gt^2

Where, h₀ = initial height

u = initial velocity

t = time

g = acceleration due to gravity

At t = 0,

Put the value into the formula

-3.053113177191196\times10^{-18}=h_{0}+0-0

h_{0}=-3.053113177191196\times10^{-18}

We need to calculate the height of ball at t = 1

Using equation of motion

h_{1}=h_{0}+u_{0}t-\dfrac{1}{2}gt^2

Put the value in the equation

6.000000000000134=-3.053113177191196\times10^{-18}+u-\dfrac{1}{2}\times32

6.000000000000134+3.053113177191196\times10^{-18}+16=u_{0}

u_{0}=22\ feet/s

Velocity is the rate of change of height with respect to time

So, velocity at 1.02 sec is given

We need to calculate the height

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

On differentiating w.r.to t

h'(t)=u-\dfrac{1}{2}g(2t)

Put the value into the formula

h'(t)=22-\times32\times(1.02)

h'(t)=-10.64\ feet/sec

Hence, The rate of change of height with respect to time is -10.64 feet/sec.

4 0
3 years ago
Suppose (for this statement only), that q is moved from the origin but is still within both the surfaces. The flux through both
o-na [289]

Answer:

-True  - True  - true   - false -false - false

Explanation:

  • True The flow depends only on the charge into the surface, not on the relative position
  • True The two vectors are radial, so their relative direction  do not changes
  • True It just depends on the charge inside
  • False, it only depends on the charge, not on the form from the integration surface
  • False, because if it has a load inside it can be considered in the center, but if the load is outside the flow lines change direction with respect to the surface
  • False The flow depends only on the load inside, not on its position
7 0
3 years ago
Particles at the very outer edge of Saturn’s A Ring are in a 7:6 orbital resonance with the moon Janus. If the orbital period of
vivado [14]

Answer:

14 hours 18 minutes.

Explanation:

ratio of number of orbits, so it completes 7 orbits in the time Janus does 6.

(16*60+41)*6/7=858 minutes or 14 hours 18 minutes

3 0
4 years ago
Solids in which the atoms have no particular order or pattern are called
posledela

Answer:

Amorphous solids are composed of atoms or molecules that are in no particular order. Each particle is in a particular spot, but the particles are in no organized pattern. Examples include rubber and wax. Crystalline solids have a very orderly, three-dimensional arrangement of atoms or molecules

Explanation:

8 0
3 years ago
Which graph accurately shows the relationship between kinetic energy and speed as speed increases?
mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = \frac{1}{2}mv^{2}

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE =  \frac{1}{2} x 2 x 3^{2}

     = 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE =  \frac{1}{2} x 2 x 4^{2}

     = 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0
3 years ago
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