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Olegator [25]
3 years ago
11

The quarterback throws a football at 25 m/s at a certain angle above the horizontal. If it took the ball 2.5 s to reach the top

of its path, how long was it in the air?
Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

The ball was in the air for 5.0 seconds

Explanation:

When the ball is thrown above at a certain angle from the horizontal it follows a projectile motion.

The ball has a specific vertical and horizontal velocity.

The time take for the ball to reach at the top of its path is also the time taken by the vertical velocity vector to become zero. This is the first half of the projectile motion (ball is going up).

For the second half, the ball comes down and it accelerates with same gravitational acceleration from which it was decelerating in the first half. Hence the time to come down will be the same 2.5 seconds.

So the total time the ball was in the air is twice of 2.5, i.e,

5.0 seconds

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What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
Wewaii [24]

The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

v = u + at

\therefore 0=\frac{2500}{9} + a(1.1)

\Rightarrow a=-\frac{2500}{9(1.1)}

\therefore a = - 252.52 \ m/s^2

Hence , the deceleration of the rocket is  252.52\ m/s^2.

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6 0
1 year ago
8 colors according to their alphabetical order​
damaskus [11]

Answer:

8 colors in alphabetical order are

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blue

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Explanation:

6 0
2 years ago
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of
natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00  m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

r'=\sqrt{d^2+r^2}

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

r'=\sqrt{(1.00)^2+(4.00)^2}

r'=\sqrt{1.00+16.00}

r'=4.12\ m

We need to calculate the path difference

Using formula of path difference

|r'-r|=4.12-4.00

|r'-r|=0.12\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Where, v = speed of sound

f = frequency

Put the value into the formula

\lambda=\dfrac{343}{300}

\lambda=1.143\ m

We need to calculate the phase difference

Using formula of phase difference

\phi=\dfrac{2\pi\times|r'-r| }{\lambda}

\phi=\dfrac{2\pi\times0.12}{1.143}

\phi=0.659\ rad

Hence, The phase difference is 0.659 rad.

7 0
3 years ago
Define in importance of building stores​
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The building sector plays a large role in the energy consumption which includes space heating or cooling, domestic hot water and electricity. Buildings with their long lifespan and huge amount of already existing buildings, makes revision in energy characteristics of a building constrained.
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