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2 years ago

A bird can fly 25 km/h. How long does it take to fly 15 km?

1 answer:

5 0

25 km/h means that 15 kilometers will be 15/25ths of an hour. Therefore, 3/5 of an hour is 36 minutes, so it would take 36 minutes to fly that distance.

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A golf ball is hit by a club. The graph shows the variation with time of the force exerted on the bal

emmasim [6.3K]

You can't find the acceleration of the ball. The graph tells the force, but you'd also need to know the mass of the ball.

5 0

2 years ago

Read 2 more answers

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

Mice21 [21]

Answer:

$v = 3369.2 m/s$

Explanation:

As we know that Beacon is rotating with angular speed

$f = 10 rev/min$

so we have

$\omega = 2\pi f$

$\omega = 2\pi(\frac{10}{60})$

$\omega = 1.047 rad/s$

now we know that

$v = r \omega$

here we will have

$r = 2 miles = 2(1609 m)$

$r = 3218 m$

so we have

$v = 3218(1.047)$

$v = 3369.2 m/s$

6 0

2 years ago

Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp

Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

8 0

2 years ago

A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

gavmur [86]

Answer : The correct option is, (d) $535^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

Now put all the given values in the above formula, we get:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

7 0

2 years ago

Read 2 more answers

Romeo lanza suavemente guijarros a la ventana de julieta y quiere que los guijarros golpeen la ventana solo con con un component

Yuki888 [10]

Answer:

$5.219\,\frac{m}{s}$

Explanation:

Las condiciones del problema requieren el cálculo de la rapidez inicial de los guijarros. Se sabe que el componente vertical de la rapidez final es cero. Por tanto, el tiempo se determina a continuación: (The conditions of this problems require the calculation of the initial speed of the peebles. It is known that vertical component of the final speed is zero. Therefore, the time is determined herein:).

$(0\,\frac{m}{s})^{2} = v_{o,y}^{2} - 2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.5\,m)$

$v_{o,y} = 9.395\,\frac{m}{s}$

$0\,\frac{m}{s} = 9.395\,\frac{m}{s} - \left(9.807\,\frac{m}{s^{2}} \right)\cdot \Delta t$

$\Delta t = 0.958\,s$

Además, se determina el componente horizontal de la rapidez inicial (Likewise, the horizontal component of the initial speed is determined):

$v_{o,x} = \frac{5\,m}{0.958\,s}$

$v_{o,x} = 5.219\,\frac{m}{s}$

El guijarro tiene una rapidez de $5.219\,\frac{m}{s}$ cuando golpea la ventana (The peeble has a speed of $5.219\,\frac{m}{s}$ when it hits the window).

6 0

2 years ago