When energy from the sun hits the air above

What are the features of image formed by convex mirror.

STatiana [176]

The image is always virtual and erect. The image is highly diminished or point sized. It is always formed between F and P.

4 0

4 years ago

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A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

Given data:

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

To find:

Final velocity: ?

Formula:

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

Solution:

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago

A weightless spring is stretched 10 cm by a suspended 1-kg block. If two such springs are used to suspend the block, one spring

leva [86]

Answer:

total stretch of the double-length spring will be 20 cm

Explanation:

given data

length x1 = 10 cm

mass = 1 kg

mass = double = 2 kg

to find out

the total stretch of the double-length spring will be

solution

we can say here spring constant is

k = mg ............1

k is spring constant and m is mass and g is acceleration due to gravity

so for in 1st case and 2nd case with 1 kg mass and 2 kg mass

kx1 = mg .........................2

and

kx2 = 2mg ........................3

x is length

so from equation 2 and 3

$\frac{kx1}{kx2}= \frac{1mg}{2mg}$

$\frac{x1}{x2} = \frac{1}{2}$

$\frac{10}{x2} = \frac{1}{2}$

x2 = 20

so total stretch of the double-length spring will be 20 cm

7 0

4 years ago

How do the meetings of the words exercise and fitness differ?

ICE Princess25 [194]

Exercise is the activity and fitness is a lifestyle and done with time

7 0

3 years ago

A small, spring-loaded cannon launches a tennis ball from level ground with an initial speed vi at an angle θi with the horizont

lora16 [44]

Answer:

a. $T\ =\ \dfrac{2v_isin\theta_i}{g}$

b. $v_icos\theta_i$

c. $v_1sin\theta_i$

Explanation:

Given,

*initial velocity of the ball = $v_i$

*angle of projection = \theta_i

Horizontal component of the initial velocity of the ball = $v_x\ =\ v_icos\theta_i$

vertical initial component of the initial velocity of the ball = $v_y\ =\ v_isin\theta_i$

part a

From the kinematics,

In the y-direction motion,

total vertical displacement of the ball during the whole motion is zero.

Ball is moving under the gravitational acceleration, therefore the acceleration of the ball = -g, because gravitational acceleration always acts in the downward direction,

Let t be the time of flight of the whole motion,

$y\ =\ v_yt\ -\ \dfrac{1}{2}gt^2\\\Rightarrow 0\ =\ v_isin\theta_i t\ -\ \dfrac{1}{2}gt^2\\\Rightarrow t\ =\ \dfrac{2v_isin\theta_i}{g}\\$

part b.

At the peak of the path of the ball, the vertical component of the velocity of the ball becomes zero, only horizontal component of the velocity acts on the ball is equal to = $v_x\ =\ v_icos\theta_i$

part c.

Initial vertical component of the velocity of the ball = $v_y\ =\ v_isin\theta_i$

5 0

3 years ago