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Slav-nsk [51]
3 years ago
14

- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7

.4 + 3.9 , and c = 7.6 + 8.8 + 2.2 . (d) Calculate the angle between r and the positive z axis. (e) What is the component of along the direction of b? (f) What is the magnitude of the component of a perpendicular to the direction of b but in the plane of a and b?
Physics
1 answer:
frez [133]3 years ago
6 0

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b=   -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+  7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

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Answer:

temperature at  326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact =  15.7-kg

specific heat of silver  cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system  is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i )  = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × 10^{5} J/kg

and we know that system is insulated

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so we can say

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )  

put here value we get

mass of silver × specific heat of silver  × ( initial temp - final temp ) = mass of  ice × specific heat of ice  × ( ice temp ) + mass of  ice × latent heat of fusion of ice + mass of  ice × specific heat of water  × (final temperature )

15.7  × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × 10^{5} + 4.6 × 4286 × ( T - 273 )

solve we get

T =  326.44 K

so temperature at  326.44 K system achieve equilibrium

7 0
3 years ago
The force exerted on the tires of a car that directly accelerate it along a road is exerted by the
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The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

<h3>What is force?</h3>

Force is defined as the product of mass and acceleration of an object.

Friction is defined as the force that resists the movement of an object over another.

Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.

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2 years ago
Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct
xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

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3 years ago
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th
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Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

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T_1=500^{\circ}F

Q=0

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We have to find the exit temperature.

By steady energy flow equation

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C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

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