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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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2 years ago
Read 2 more answers
Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o
spayn [35]

To solve letter a:

d1 = 85t1 = 16 km, 
85t1 = 16, 
t1 = 16 / 85 = 0.1882 h = 11.29 min. 

d2 = 115t2 = 16 km, 
115t2 = 16, 
t2 = 16 / 115 = 0.139 h = 8.35 min. 

t1 - t2 = 11.29 - 8.35 = 2.94 min. 
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h. 
d1 = d2, 
115t = 85(t + 0.25), 
115t = 85t + 21.25, 
115t - 85t = 21.25, 
30t = 21.25, 
t = 21.25 / 30 = 0.71 h, 

d = 115 * 0.71 = 81.65 km.

8 0
3 years ago
I need the ans for this question QUICK PLEASE!!!
Basile [38]

Explanation:

A) 1.05

B) 1.33

C) 1.16

D) 0.62

All units in cm

3 0
3 years ago
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i
gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

I = 4.2 kg × 0.0626 m²

I = 0.2625 kg.m²

So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

∝ = T / I

we substitute

∝ = 4.104 / 0.2625

∝ = 15.65 rad/s²

Therefore, the resulting angular acceleration is 15.65 rad/s²

8 0
3 years ago
A 2320 pound roller coaster starts from rest and is launched such that it creates a 110 ft high hill with a speed of 65 mph. The
Sergeu [11.5K]

Answer: E = 941738.537J

Explanation:

to begin,

given that the mass = 2320 pound = 1052.334 kg

Δh = 110 ft = 33.528 m

given that  distance (d) = 1283 ft = 391.058 m

also the speed (v) is 65 mph = 29.058 m/s

force (F) = 87 pounds = 386.995 N

from our knowledge in work energy theory;

E = Fd + 1/2mv² + mgh

E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)

E = 151337.491 + 444278.2 + 346122.84

E = 941738.537J

i hope this helps, cheers.

3 0
3 years ago
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