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Licemer1 [7]
3 years ago
12

An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 5

0, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Physics
1 answer:
rosijanka [135]3 years ago
5 0

Answer:

The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

Explanation:

Given that,

Mass of object = 2.5 kg

Momentum p= 3i+6j+7k

Force F=50i+50j+100k

Time t=5\times10^{-3}\ s

We need to calculate the momentum of the object at the end

Using formula of impulse

J=\Delta p

J=m\Delta v...(I)

J=F\times\Delta t....(II)

From equation (I) and (II)

F\times\Delta t=m\Delta v

F\times \Delta t=m(v_{f}-v_{i})

mv_{f}=F\times \Delta t+mv_{i}

Put the value into the formula

p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k

p_{f}=0.25i+0.25j+0.5k+3i+6j+7k

p_{f}=(3.25i+6.25j+7.5k)\ kg m/s

Hence, The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

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61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A
Mademuasel [1]

Answer:

Part a)

percentage = 21.3%

Part b)

percentage = 2.13 \times 10^{-5}%

Explanation:

As we know that total power used in the room is given as

P = P_1 + P_2 + P_3 + P_4

here we have

P_1 = (110)(3) = 330 W

P_2 = 100 W

P_3 = 60 W

P_4 = 3 W

P = 330 + 100 + 60 + 3

P = 493 W

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

110\times i = 493

i = 4.48 A

now resistance of transmission line

R = \frac{\rho L}{A}

R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}

R = 5.23 \ohm

now power loss in line is given as

P = i^2 R

P = (4.48)^2(5.23)

P = 105 W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{105}{493} \times 100

percentage = 21.3%

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

110 \times 10^3 (i ) = 493

i = 4.48\times 10^{-3} A

now power loss in line is given as

P = i^2 R

P = (4.48 \times 10^{-3})^2(5.23)

P = 1.05 \times 10^{-4} W

Now percentage loss is given as

percentage = \frac{loss}{supply} \times 100

percentage = \frac{1.05 \times 10^{-4}}{493} \times 100

percentage = 2.13 \times 10^{-5}%

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3 years ago
A pilot can withstand an acceleration of up to 9g, which is about 88 m/s2, before blacking out. What is the acceleration experie
Harman [31]

Answer:

Explanation:

Given that, the pilot can withstand 9g acceleration which is approximately 88m/s².

Now, the pilot is traveling in a circle of radius

r = 3340 m

And the speed is

v = 495 m/s

Then, acceleration?

The acceleration of a circular motion can be determine using centripetal acceleration

a = v² / r

a = 495² / 3340

a = 73.36 m/s².

Since the acceleration is less that the acceleration the pilot can withstand, then, I think the pilot makes the turn without blacking out and successfully

4 0
3 years ago
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