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Advocard [28]
3 years ago
14

Vector A has magnitude of 8units and makes an angle of 45° with the positive x-axis. Vector B also has the same magnitude of 8un

its and directed along the negative x-axis. Find a. The magnitude and direction of A+B b. The magnitude and direction of A-B​
Physics
1 answer:
iragen [17]3 years ago
3 0

Answer:

Explanation:

Because vectors have direction and x and y components you can't just add them and say that their length is 16 because A is 8 units and so is B. What you're actually finding is the magnitude and direction of the vector that results from this addition. The magnitude is the length of the resultant vector, which comes from the x and y components of A and B, and the direction is the angle between the resultant vector and the positive x axis. To add the vectors, then, we need to find the x and y components of each. We'll do the x components of A and B first so we can add them to get the x component of C. Since x values are directly related to cos, the formula to find the x components of vectors is

V_x=Vcos\theta which is the magnitude of the vector (its length) and the angle. Finding the x components of A:

A_x=8.0cos45 so

A_x=5.7 and for B:

B_x=8.0cos180 since the negative x axis is the 180 degree axis and

B_x=-8.0 If we add them, we get

C_x=-2.3

Now onto the y components. The formula for that is almost the same as the x components except use sin instead of cos:

A_y=8.0sin45 so

A_y=5.7 and

B_y=8.0sin180 so

B_y=0 If we add them, we get

C_y=5.7

Now for the final magnitude:

C_{mag}=\sqrt{(-2.3)^2+(5.7)^2} and

C_{mag}=6.1 units and now onto the direction.

The x component of C is positive and the y component is negative, which means that the direction has us at an angle is quadrant 2; we add 180 to whatever the angle is. Finding the angle:

tan^{-1}(\frac{C_y}{C_x})=(\frac{5.7}{-2.3}) = -68 + 180 = 112 degrees

The resultant vector of A + B has a magnitude of 6.1 and a direction of 112°

Do the same thing for subtraction, except if you're subtracting B from A, the direction that B is pointing has to go the opposite way. That means that A doesn't change anything at all, but B is now pointing towards 0.

A_x=5.7 (doesn't change from above)

B_x=8.0cos0 and

B_x=8.0 so

C_x=13.7 and

A_y=5.7 (also doesn't change from above)

B_y=8.0sin0 so

B_y=0 and

C_y=5.7 and for the magnitude:

C_{mag}=\sqrt{(13.7)^2+(5.7)^2 so

C_{mag}=15units and for the direction:

tan^{-1}(\frac{5.7}{13.7})=23 and since both x and y components of C are in Q1, we add nothing.

And you're done!!!

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The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

\Delta T=\frac{Q}{mC_s}

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For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

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Therefore, by re-arranging the equation we find the mercury's specific heat:

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Learn more about specific heat capacity:

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ira [324]

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3 years ago
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr
KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s

Time the parachutist falls without friction is 3.19 seconds

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2

s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2

Now the initial velocity of the last half height will be the final velocity of the first half height.

v=u+at\\\Rightarrow v=at

Since the height are equal

\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0

t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2

Magnitude of acceleration is -43200 ft/s²

5 0
3 years ago
During a tennis serve, a racket is given an angular acceleration of magnitude 150 rad/s^2. At the top of the serve, the racket h
QveST [7]

Answer:

270 m/s²

Explanation:

Given:

α = 150 rad/s²

ω = 12.0 rad/s

r = 1.30 m

Find:

a

The acceleration will have two components: a radial component and a tangential component.

The tangential component is:

at = αr

at = (150 rad/s²)(1.30 m)

at = 195 m/s²

The radial component is:

ar = v² / r

ar = ω² r

ar = (12.0 rad/s)² (1.30 m)

ar = 187.2 m/s²

So the magnitude of the total acceleration is:

a² = at² + ar²

a² = (195 m/s²)² + (187.2 m/s²)²

a = 270 m/s²

3 0
3 years ago
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