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Nikitich [7]
3 years ago
10

What measurements or observations tells you that a car is accelerating

Physics
2 answers:
lutik1710 [3]3 years ago
8 0

You need to observe the car at two different times.

-- The first time: 
You write down the car's speed, and the direction it's pointing.

-- The second time:
You write down the car's speed and the direction it's pointing, again.

You take the data back to your lab to analyze it.

-- You compare the first and second speed. If they're different,
then the car had acceleration during the time between the two
observations.

-- You compare the first and second direction.  If those are different,
even if the speeds are the same, then the car had acceleration during
the time between the two observations.

(Remember, "acceleration" doesn't mean "speeding up". 
It means any change in speed or direction of motion.)
PtichkaEL [24]3 years ago
7 0
<span>Frame of reference. It</span> uses the second law of motion to determine whether it is moving or not...
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It may seem strange that the selected velocity does not depend on either the mass or the charge of the particle. (For example, w
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Answer:

b) q large and m small

Explanation:

q is large and m is small

We'll express it as :

q > m

As we know the formula:

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Assume that you want a velocity selector that will allow particles of velocity v⃗  to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.

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Atoms with the same number of protons but with different electrical charges _____.
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Read 2 more answers
A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of
Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

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

h=(u\sin60)\times t-\dfrac{1}{2}gt^2

Put the value into the formula

h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2

h=41.6\ m

(b). We need to calculate the maximum height of the ball

Using formula of height

h_{max}=\dfrac{(u\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\sin60)^2}{2\times 9.8}

h=41.67\ m

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

v=u-gt

v=u\sin\theta-gt

Put the value into the formula

v_{y}=33\times\sin 60-9.8\times3.0

v_{y}=-0.82\ m/s

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=33\times\cos60

v_{x}=16.5\ m/s

We need to calculate the ball's impact speed

Using formula of velocity

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(16.5)^2+(-0.82)^2}

v=16.52\ m/s

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0
2 years ago
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