All categories

3 years ago

Please answer the question in the picture, worth 25 points

Physics

1 answer:

viktelen [127]3 years ago

5 0

Answer:

Option C, increases and decreases

Explanation:

When an object making noise approaches you, the wave frequency increases leading to a higher pitch. Conversely, when it moves away from you or retreats, the wave frequency decreases leading to a lower pitch. This can be observed in ambulance sirens.

You might be interested in

If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?

goldfiish [28.3K]

An energy bar contains of 20grams of carbohydrates and 1gram is equal to 17 KJ. If the energy bar was his only fuel the total energy available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and his Distance is equal to (5,000 x .2209) = 1.1045km.

3 0

3 years ago

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

A gardener pushes a wheelbarrow around the edges of a lawn. When she has finished, she is 5 m from her starting point. She has t

zvonat [6]

Answer:

Work done by the gardner is 500 J

Explanation:

As we know that the gardner apply force perpendicular upward by magnitude 300 N and along the floor horizontal force is 100 N

so we have

$F = 100 \hat i + 300 \hat j$

now the displacement of the gardner along the floor is

$d = 5\hat i$

now work done is given as

$W = F. d$

so we have

$W = (100 \hat i + 300 \hat j). (5\hat i)$

$W = 500 J$

3 0

3 years ago

Calculate the number of moles in 100g of water

dsp73

5.55 mol H2O

Explanation:

Water has a molar mass of 18.01528 g/mol. We can then calculate the number of moles of water as

100 g H20 × (1 mol H2O/18.01528 g H20)

= 5.55 mol H2O

5 0

3 years ago

A little girl is walking, swinging her arms through a 20° angle every 0.25 s. The length of each arm is 30 cm. Assuming that the

rodikova [14]

Answer:

The force will be " $9.8\times 10^{-3} \ N$ ".

Explanation:

The given values are:

Mass,

m = 1 gram

Angle,

Ф = 20°

As we know,

⇒ $F=mg$

On substituting the given values in the above expression, we get

⇒ $=(1.0\times 10^{-3})(9.8)$

⇒ $=9.8\times 10^{-3} \ N$

3 0

2 years ago