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1 year ago

If a ball was thrown upward at 46.3 m/s how long would the ball stay in the air

1 answer:

8 0

$V = Vo + a.t

$

The ball is against the vector of gravity. Then, the gravity will be negative.

$0 = 46,3 + (-9.8).t \\ 
t = \frac{46.3}{9.8} \\ 
t \approx 4.72 

$

The ball will stop in the air after approx. 4.72 seconds. And will take the same time to hit the ground.

It will stay approx. 9.44 seconds in the air.

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A shear wave (S wave) is a type of seismic _________ that shakes the ground back and forth perpendicular to the direction the wa

Zepler [3.9K]

Body waves

Explanation:

A shear wave(S-wave) is a type of seismic body waves that shakes the ground back and forth perpendicular to the direction the wave is moving.

Seismic waves are elastic waves usually generated when there is a disturbance within the earth.
There are two types of seismic waves:

Surface waves

Body waves

Body waves travel within the earth and they cause disturbances there. P and S waves are the two types of body waves that we have.
Surface waves travels on the earth surface. They are the love and rayleigh waves. They are the ones that cause destruction on the earth surface during an earthquake.

Learn more:

Earthquake brainly.com/question/6520403

#learnwithBrainly

7 0

2 years ago

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago

The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Marat540 [252]

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

7 0

2 years ago

Read 2 more answers

A force of 200 N is applied on small piston of a pascal press. What would be the

VladimirAG [237]

Answer:

The force applied on the big piston is 1306.67 N

Explanation:

Given;

force applied on small piston, F₁ = 200 N

diameter of the small piston, d₁ = 4.37 cm

radius of the small piston, r₁ = d₁/2 = 2.185 cm