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Arada [10]
3 years ago
14

Analyze at the image below and answer the question that follows.

Physics
2 answers:
Rudiy273 years ago
8 0

the correct answer is c.

Rufina [12.5K]3 years ago
3 0

Answer:

C

Explanation:

correct on edgenuity

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What does your body do with nutrients during the digestion process?
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Answer:

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Explanation:

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6 0
3 years ago
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring t
dmitriy555 [2]

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

7 0
3 years ago
When you are in the front passenger seat of a car turning to the left, you may find yourself pressed against the right-side door
Anna71 [15]
When you are in the front passenger seat of a car turning to the left, you may find yourself pressed against the right-side door. because of <span> Newton's first and third law</span>
6 0
3 years ago
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borishaifa [10]

Answer:

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Explanation:

8 0
3 years ago
A wheel of radius R = 0.80 m is pulled by a rope looped around a frictionless axle. The total mass of the wheel and axle assembl
iragen [17]

Answer:

\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18

Explanation:

To find the rotational kinetic energy you first calculate the angular acceleration by using the following formula:

\tau=I\alpha\\\\FR=I\alpha\\\\\alpha=\frac{FR}{I}

F: force applied

R: radius of the wheel

I: moment of inertia

\alpha=\frac{(120N)(0.80m)}{26.88kgm^2}=3.57\frac{rad}{s^2}

With this value you calculate the angular velocity:

\omega^2=\omega_o^2+2\alpha \theta\\

you calculate how many radians the wheel run in 5.0m

\theta=\frac{2\pi (0.8m)}{5.0m}=\approx1.00rad

\omega=\sqrt{2(3.57\frac{rad}{s^2})(1.00rad)}=2.67\frac{rad}{s}

Next, you use the formula for the rotational kinetic energy:

K_R=\frac{1}{2}I\omega^2=(26.88)(2.67)^2 J = 188.63J

For the transnational kinetic energy you use the following equation:

W=\Delta K_T  (net work equals the change in the kinetic energy).

By replacing the you obtain:

\Delta K_T=Fd=(120)(5.0)J=600J

Finally, the ratio between translational rotational kinetic energy is:

\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18

hence, translational kinetic energy is three times the rotational kinetic energy.

6 0
3 years ago
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