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iren [92.7K]
2 years ago
9

Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of north.

The resultant vector A + B is given by:
(a) 9.8 m at an angle of 26° north of east,


(b) 3.3 m at an angle of 26° north of east,


(c) 7.2 m at an angle of 26° east of north,


(d) 3.3 m at an angle of 64° east of north or


(e) 9.8 m at an angle of 64° east of north
Physics
1 answer:
sladkih [1.3K]2 years ago
3 0

Answer:

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

C

Explanation:

Resolving the vectors to vertical and horizontal component;

Vertical;

Vector A = 6sin30

Vector B = 4sin60

Resultant vertical = 6sin30 + 4sin60 = 6.464m

Horizontal;

Vector A = 6cos30

Vector B = -4cos60

Resultant horizontal = 6cos30 - 4cos60 = 3.196m

Resultant R = √(6.464^2 + 3.196^2) = 7.2m

Tanθ = 6.464/3.196

θ = taninverse (6.464/3.196) BN

θ = 64° north of East.

Or

26° east of north

The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,

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Answer:

distance=6.11m

Explanation:

A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How far does the player run before he catches the ball? Ignore air resistance. I got stuck because I wasn't sure which formula to use when approaching this problem. Does it involve an angle at all?

first of all we get the time it takes to reach the maximum height

then twice of the time it takes to reach maximum height will be the time of flight

from newtons equation of motion

v=u+at

v=0

u=6m/s

0=6-9.81t

t=.61s

the time of flight will be 1.22secs

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distance=speed*time

distance=2.5m/s*1.22

distance=6.11m

3 0
2 years ago
A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is
vova2212 [387]

solution:

consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
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2 years ago
Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy
bekas [8.4K]
The energy of a photon is given by:
E=hf
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon, E=3.26 \cdot 10^{-19} J, so we can rearrange the equation to calculate the frequency of the photon:
f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz

And now we can use the following relationship between frequency f, wavelength \lambda and speed of light c to find the wavelength of the photon:
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An earth satellite in an elliptical orbit travels fastest when it is
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ANY orbiting object travels fastest when it's closest to the central body.
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In t-ball, young players use a bat to hit a stationary ball off a stand. The 140 g ball has about the same mass as a baseball, b
fomenos

Question is missing. Found on google:

<em>"Part A What is the acceleration of the ball? Express your answer to two significant figures and include the appropriate units. </em>

<em>Part B </em>

<em>What is the net force on the ball during the hit?  </em>

<em>Express your answer to two significant figures and include the appropriate units."</em>

Solution:

A) 6000 m/s^2

The acceleration of the ball is given by

a=\frac{v-u}{t}

where

v = 12 m/s is the final velocity

u = 0 is the initial velocity (the ball is stationary)

t = 2.0 ms = 0.002 s is the time of contact

Substituting,

a=\frac{12-0}{0.002}=6.0 \cdot 10^3 m/s^2

B) 8.4\cdot 10^2 N

The force on the ball can be found by using Newton's second law:

F=ma

where

m = 140 g = 0.14 kg is the mass of the ball

a=6.0\cdot 10^3 m/s^2 is the acceleration

Substituting,

F=(0.14)(6.0\cdot 10^3)=8.4\cdot 10^2 N

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2 years ago
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