All categories

2 years ago

PLEASE HELP :)

1 answer:

6 0

1) in the opposite direction always
2) drag force is the force that acts in the opposite direction of the applied force (air resistance if a drag force for example)
3) frictions forces between the sole and the ground as the owner walks
4) chalk would glide very easily, so a creaking sound could never be produced, but also one could not write anything on the board!
5) I apologize for my lack of familiarity with the kind of sport but the probable answer is to raise the friction of their hands in order to decrease slipping and get a tighter, more firm grip

You might be interested in

A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

2 years ago

(b) During one day, 250 kg of water is pumped through

ser-zykov [4K]

Answer: The energy incident on the solar panel during that day is $9.24 \times 10^{7} J$ .

Explanation:

Given: Mass = 250 kg

Initial temperature = $16^{o}C$

Final temperature = $38^{o}C$

Specific heat capacity = 4200 $J/kg^{o}C$

Formula used to calculate the energy is as follows.

$q = m \times C \times (T_{2} - T_{1})$

where,

q = heat energy

m = mass of substance

C = specific heat capacity

$T_{1}$ = initial temperature

$T_{2}$ = final temperature

Substitute the values into above formula as follows.

$q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C$

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

$\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J$