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erica [24]
3 years ago
12

Find the components to write this vector in unit vector notation: 63.5 A ​please help

Physics
1 answer:
IrinaVladis [17]3 years ago
6 0

Vector is perpendicular to x axis or i component.

Hence i component is 0

j component is 63.5

\\ \sf\longmapsto \overrightharpoon{A}=0\hat{i} +63.5\hat{j}

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Answer:

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Explanation:

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2 years ago
Help fast please due tomorrow
Nitella [24]

Answer:

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Explanation:

8 0
2 years ago
For a freely falling object weighing 3 kg : A. what is the object's velocity 2 s after it's release. B. What is the kinetic ener
Fed [463]

A) 19.6 m/s (downward)

B) 576 J

C) 19.6 m

D) Velocity: not affected, kinetic energy: doubles, distance: not affected

Explanation:

A)

An object in free fall is acted upon one force only, which is the force of gravity.

Therefore, the motion of an object in free fall is a uniformly accelerated motion (constant acceleration). Therefore, we can find its velocity by applying the following suvat equation:

v=u+at

where:

v is the velocity at time t

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration due to gravity

For the object in this problem, taking downward as positive direction, we have:

u=0 (the object starts from rest)

a=9.8 m/s^2

Therefore, the velocity after

t = 2 s

is:

v=0+(9.8)(2)=19.6 m/s (downward)

B)

The kinetic energy of an object is the energy possessed by the object due to its motion.

It can be calculated using the equation:

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is the speed of the object

For the object in the problem, at t = 2 s, we have:

m = 3 kg (mass of the object)

v = 19.6 m/s (speed of the object)

Therefore, its kinetic energy is:

KE=\frac{1}{2}(3)(19.6)^2=576 J

C)

In order to find how far the object has fallen, we can use another suvat equation for uniformly accelerated motion:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the object in free fall in this problem, we have:

u = 0 (it starts from rest)

a=g=9.8 m/s^2 (acceleration of gravity)

t = 2 s (time)

Therefore, the distance covered is

s=0+\frac{1}{2}(9.8)(2)^2=19.6 m

D)

Here the mass of the object has been doubled, so now it is

M = 6 kg

For part A) (final velocity of the object), we notice that the equation that we use to find the velocity does not depend at all on the mass of the object. This means that the value of the final velocity is not affected.

For part B) (kinetic energy), we notice that the kinetic energy depends on the mass, so in this case this value has changed.

The new kinetic energy is

KE'=\frac{1}{2}Mv^2

where

M = 6 kg is the new mass

v = 19.6 m/s is the speed

Substituting,

KE'=\frac{1}{2}(6)(19.6)^2=1152 J

And we see that this value is twice the value calculated in part A: so, the kinetic energy has doubled.

Finally, for part c) (distance covered), we see that its equation does not depend on the mass, therefore this value is not affected.

5 0
2 years ago
A residential subdivision encompasses 1100 acres with a housing density of four houses per acre. Assume that a high-value reside
aleksandrvk [35]

Answer:

(a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

Explanation:

Given that,

Area = 1100 acres

Number of house in 1 acres = 4

\text{Number of house in 1100 acres} = 4\times1100

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Per house water demand = 800 g/day/house

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\text{average daily demand}=house\times\text{Per house water demand}

\text{average daily demand}=4400\times800\ gallon/day

\text{average daily demand}=3520000\ gallon/day

\text{average daily demand}=\dfrac{3520000}{24\times60}\ gallon/min

\text{average daily demand}=2444.44\ gallon/min

The average daily demand of this subdivision is 2444.44 gallon/min.

(b). We need to calculate the design-demand used to design the distribution system

Using formula for the design-demand

\text{design demand}=(Q_{max})daily\times\text{fire flow}

\text{design demand}=1.64\times2444.4\times1000

\text{design demand}=4008816\ gallon/m

Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.

(b). The design-demand used to design the distribution system is 2444.44 gallon/min.

8 0
2 years ago
<img src="https://tex.z-dn.net/?f=what%20%5C%3A%20is%20%5C%3A%20unit%20%5C%3A%20of%20%5C%3A%20pressure%20%5C%3A%20%20%7B%3F%7D%2
8_murik_8 [283]

Answer:

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hope it is helpful to you

5 0
3 years ago
Read 2 more answers
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