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3 years ago

Which of the following statements best explains why a book resting on a table is in equilibrium?

1 answer:

3 0

I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

You might be interested in

A snail at position 3 cm moves to position 20 cm in 8 seconds.

natulia [17]

Answer: 17cm.

Explanation:

The equation you're using is:

Δd = df - di

Which means the change in position is equal to the final position minus the starting position. In this case that works out to 20cm - 3cm = 17cm. We're only interested in how much the snail moved, not how long it took to move, so even though they give a time it actually doesn't matter for this question.

4 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago

Pressure is always measured in comparison to a standard pressure. True False

irakobra [83]

Answer:

true

Explanation:

8 0

3 years ago

At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

6 0

3 years ago

An object with a mass of 0.5 kilometre start from rest and achieves a maximum speed of 20 metre per second in 0.01 second, what

Katarina [22]

Answer:

Hiii how are you doing??I don't understand that

3 0

2 years ago