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lesya [120]
3 years ago
12

What is the y component of a vector defined as 12.2m at 81.5°?

Physics
1 answer:
sergejj [24]3 years ago
3 0

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

V_y=Vsin\theta which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

V_y=12.2sin(81.5) and get

V_y=12.1 m

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In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he
Rzqust [24]

Answer:

Increasing until terminal velocity is reached

Explanation:

Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.

5 0
3 years ago
What idea did Max Planck propose to help explain why a blackbody radiator did not give off light of increasingly high frequency
Leno4ka [110]
The idea that <span>Max Planck propose to help explain why a blackbody radiator did not give off light of increasingly high frequency as its temperature increased is that </span>C. Matter can absorb light only in certain specific amounts. 
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4 years ago
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A object with a mass of 1.5 kg is lifted from the ground to a height of 0.22 m what is the objects potential energy
svet-max [94.6K]

Answer:

<h2>3.3 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 1.5 × 10 × 0.22

We have the final answer as

<h3>3.3 J</h3>

Hope this helps you

5 0
3 years ago
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use
saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

M = \frac{Nl}{f_ef_0}

Where

N = Near point

l = distance between the object lens and eye lens

f_0= Focal length

f_e= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

M = \frac{25*10}{3*5}

M = 16.67\approx 17\\

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3 0
3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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