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Nana76 [90]
3 years ago
11

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push

es back against the motion.
What is the acceleration of the cart?
Physics
1 answer:
atroni [7]3 years ago
3 0

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N      Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N    mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

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An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does h
Rainbow [258]

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The moment of inertia decreased by a factor of 4

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Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

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During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

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I₂ is the final moment of inertia

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4 0
3 years ago
two ships leave a port at the same time. The first ship sails on a bearing of 40 degrees at 18 knots and the second at a bearing
Yuliya22 [10]

Answer:

The distance between the ships is 87.84 km.

Explanation:

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Speed of first ship = 18 knots

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Speed of second ship = 26 knots

We need to calculate the resultant velocity

Using cosine rule

v=\sqrt{v_{1}^2+v_{2}^2-2v_{1}v_{2}\cos\theta}

Put the value into the formula

v=\sqrt{18^2+26^2-2\times18\times26\times\cos90}

v=\sqrt{18^2+26^2}

v=\sqrt{324+676}

v=10\sqrt{10}

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d =v\times t

Put the value into the formula

d=10\sqrt{10}\times1.5\times1.852

d=87.84\ km

Hence, The distance between the ships is 87.84 km.

7 0
3 years ago
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