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2 years ago

Calculate the speed of a car that traveled 200 kilometers in 3 hours.

Physics

2 answers:

nignag [31]2 years ago

8 0

Answer:

The units (km/h) tell you how to do this! 200km/3h = 66.66666666…. BUT technically you only have ONE significant digit: 3 so 66.666… rounded to ONE digit is 70km/h but that is probably not important in this intro class so V = 66.67 or 67 km/h

Kamila [148]2 years ago

7 0

Answer:

66.7km/h

Explanation:

s=d/t

200/3

=66.6666666

round off: 66.7km/h

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What is inertia and how is it related to the newton's first law of motion?

Alina [70]

Answer:

Inertia is the resistance of any physical object to any change in its velocity. This includes changes to the object's speed, or direction of motion. An aspect of this property is the tendency of objects to keep moving in a straight line at a constant speed, when no forces act upon them.

Explanation:

Some sort of a local field, maybe not our A field, is really the cause of inertia. When you push on an object a gravitational disturbance goes propagating off into either the past or the future. Out there in the past or future the disturbance makes the distant matter in the universe wiggle.

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2 years ago

A cobalt atom contains 27 protons, 32 neutrons, and 27 electrons. What is

kirza4 [7]

32? I could be wrong but I’m going with that answer choice

3 0

2 years ago

The graph represents velocity over time...

Nady [450]

Answer:

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8 0

2 years ago

Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug

Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

$F_D=\dfrac{1}{2}\rho C_DA v^2$

$C_D$ =Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

Now by putting the values

$\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}$

$\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}$

$\dfrac{F_D_1}{F_D_2}=0.444$

Percentage Change in the drag force

$\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100$

$\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100$

$\Delta F=\dfrac{1-0.444}{0.444}\times 100$

ΔF=125.22 %

Therefore force will increase by 125.22 %.

3 0

3 years ago

A mass m is attached to an ideal massless spring with spring constant k. In experiment 1 the mass oscillates with amplitude a, a

Trava [24]

Time period remains the same in both the experiment as change in amplitude does not affect time period.

What are the factors on which time period depends in SHM?

Time period is given by:

$T=2\pi \sqrt{\frac{m}{k} }$

where,

T = time period

m = mass

k = spring constant

In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.

What is Spring Constant?

A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.

Learn more about SHM here:

brainly.com/question/20885248

#SPJ4

6 0

8 months ago