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3 years ago

I will mark brainlist please help

Physics

1 answer:

const2013 [10]3 years ago

8 0

Answer:

1. c) Mass

2. D) 1/2

go for it!!

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3 years ago

While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000

Minchanka [31]

Answer:

The hiker (you ) is 200 m below his/her(your) starting point

The resultant displacement in the north east direction is

$a = 6562.0 \ m$

The resultant displacement in vertical direction (i.e the altitude change )

$b =6503.1 \ m$

Explanation:

From the question we are told that

The displacement in the morning is $S_{morning} = (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)$

The displacement in the afternoon is $S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)$

Generally the direction west is negative , the direction east is positive

the direction south is negative , the direction north is positive

resultant displacement is mathematically evaluated as

$(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east$

$(4000\ m\ north) + (2500 \ m ,\ north) = 6500 \ m ,\ north$

$(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m$

From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

$a = \sqrt{900^2 + 6500^2}$

=> $a = 6562.0 \ m$

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

$b = \sqrt{6500^2 +(-200)^2 }$

=> $b =6503.1 \ m$

5 0

4 years ago

A baseball is thrown at an angle of 20° relative to the ground at a speed of 25 m/s if the ball was caught 50 m from the thrower

scoray [572]

Answer:

2.1 s

Explanation:

The motion of the ball is a projectile motion. We know that the horizontal range of the ball is

$d = 50 m$

And that the initial speed of the ball is

$u=25 m/s$

at an angle of

$\theta=20^{\circ}$

So, the horizontal speed of the ball (which is constant during the entire motion) is

$u_x = u cos \theta = 25 \cdot cos 20^{\circ} = 23.5 m/s$

And since the horizontal range is 50 m, the time taken for the ball to cover this distance was

$t=\frac{d}{u_x}=\frac{50}{23.5}=2.1 s$

which is the time the ball spent in air.

8 0

3 years ago