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GenaCL600 [577]
3 years ago
12

Please help on this one?

Physics
1 answer:
Valentin [98]3 years ago
4 0

I would say C as a primary source is a first person encounter meaning the information given should be in detail.

D is incorrect as we cannot infer a scientist written it

B is incorrect because the level of reading for a person does not depend on what type of source it is from

A would not be correct as a secondary source is a person writing about someone else's work

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A person drives a car around a circular road with a constant speed of 20 m/s. The
ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0
3 years ago
Read 2 more answers
A car initially traveling at 27.2 m/s undergoes a constant negative acceleration of magnitude 1.90 m/s2 after its brakes are app
zubka84 [21]

Answer:

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

Explanation:

We can use the following equation:

\omega_{f}^{2}=\omega_{i}^{2}-2\alpha \Delta \theta (1)

The angular acceleration is:

a_{tan}=\alpha R

\alpha=\frac{1.9}{0.325}

\alpha=5.85\: rad/s^{2}

and the initial angular velocity is:

\omega_{i}=\frac{v}{R}

\omega_{i}=\frac{27.2}{0.325}

\omega_{i}=83.69\: rad/s

Now, using equation (1) we can find the revolutions of the tire.

0=83.69^{2}-2*25.85 \Delta \theta

\Delta \theta=135.47\: rad

Therefore, the revolutions that each tire makes is:

\Delta \theta=22\: rev

I hope it helps you!

6 0
3 years ago
What is the car's average velocity (in m/s) in the interval between t = 1.0 s<br> to t = 1.5 s?
natali 33 [55]

Answer:

1.4 m/s

Explanation:

From the question given above, we obtained the following data:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Velocity (v) =..?

The velocity of an object can be defined as the rate of change of the displacement of the object with time. Mathematically, it can be expressed as follow:

Velocity = change of displacement /time

v = Δd / Δt

Thus, with the above formula, we can obtain the velocity of the car as follow:

Initial Displacement (d1) = 0.9 m

Final Displacement (d2) = 1.6 m

Change in displacement (Δd) = d2 – d1 = 1.6 – 0.9

= 0.7 m

Initial time (t1) = 1.5 secs

Final time (t2) = 2 secs

Change in time (Δt) = t2 – t1

= 2 – 1.5

= 0.5 s

Velocity (v) =..?

v = Δd / Δt

v = 0.7/0.5

v = 1.4 m/s

Therefore, the velocity of the car is 1.4 m/s

4 0
3 years ago
When a driver presses the brake pedal, his car can stop with an acceleration of -5.4m/s^2. How far will the car travel while com
Dahasolnce [82]
Information that is given:
a = -5.4m/s^2
v0 = 25 m/s
---------------------
S = ?
Calculate the S(distance car traveled) with the formula for velocity of decelerated motion:
v^2 = v0^2 - 2aS
The velocity at the end of the motion equals zero (0) because the car stops, so v=0.
0 = v0^2 - 2aS
v0^2 = 2aS
S = v0^2/2a
S = (25 m/s)^2/(2×5.4 m/s^2)
S = (25 m/s)^2/(10.8 m/s^2)
S = (625 m^2/s^2)/(10.8 m/s^2)
S = 57.87 m
3 0
3 years ago
How high would you have to lift a 1000kg car to give it a potential energy of:
Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height  = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

 It is mathematically expressed as:

          P.E  = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

               h = \frac{P.E }{mg}

<u>For 2.0 x 10³ J;</u>

                  h  = \frac{2000}{1000 x 9.8}   = 0.204m

<u>For 2.0 x 10⁵ J;</u>

                  h  = \frac{200000}{9.8 x 1000}   = 20.4m

<u>For 1.0kJ  = 1 x 10³J; </u>

                  h  = \frac{1000}{9.8 x 1000}   = 0.102m

   

5 0
3 years ago
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