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myrzilka [38]
3 years ago
12

if a 3-kg object has a momentum of 33 kg��m/s, what's its velocity? a. 99 m/s b. 36 m/s c. 11 m/s d. 9.1 m/s

Physics
2 answers:
Lubov Fominskaja [6]3 years ago
8 0

Answer : The correct option is, (c) 11 m/s

Solution : Given,

Momentum of an object = 33 Kgm/s

Mass of an object = 3 Kg

Momentum : It is the product of the mass and the velocity of an object.

Formula used :

p=m\times v

where,

p = momentum of an object

m = mass of an object

v = velocity of an object

Now put all the given values in the above formula, we get the velocity of an object.

33\text{ Kg }m/s=(3Kg)\times v

v=11m/s

Therefore, the velocity of an object is, 11 m/s

horsena [70]3 years ago
6 0
M=3kg
p=33kg.m/s
p=m*v
v=p/m
=33/3
=11m/s
thus option (c)
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A 2 kg object with a weight of 20 N is being pulled up by a rope with a tension of 12N what is the acceleration of the object
son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

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tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2 (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

7 0
3 years ago
Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr
natima [27]
<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.  

It is given by

τ = I x α          -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em>=> ω = 7.29 × 10⁻⁵ rad/s</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = \frac{1}{2} x I x ω

E = \frac{1}{2} x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

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Answer:

Fnet - Fg

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The weight, W, of an object can be expressed as;

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If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

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Apparent weight (ma) = W_{net} - mg

3 0
2 years ago
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