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2 years ago

I need help for a assignment NOT A QUIZ

Physics

2 answers:

fomenos2 years ago

7 0

Answer:

Its C

Explanation:

Debora [2.8K]2 years ago

3 0

Answer:

what

ITS BLANK FOR ME I WISH I COULD HELP YOU

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Given: Saturated air changes temperature by 0.5°C/100 m. The air is completely saturated at the dew point. The dew point has bee

erma4kov [3.2K]

Answer:

Explanation:

Given

saturated air temperature by $0.5^{\circ}C/100 m$

Dew point temperature is given by $t=2^{\circ}C$

Dew point is defined as the temperature after which air no longer to uphold the water vapor fuse with it and some water vapor may condense to a liquid.

air continues to rise for 1400 m

i.e. change in temperature would be $\Delta t =\frac{0.5}{100}\times 1400=7^{\circ}C$

Final temperature $t_f$

$t_f+\Delta t=t$

$t_f=2-7=-5^{\circ}C$

3 0

3 years ago

An outside thermometer reads 57°F. What is this temperature in °C? Round your answer to the nearest whole number.

marishachu [46]

Answer: It'd be 14.

Explanation:

The formula for this equation would be (57f-32)×5/9 which is equal to 13.889; and rounding that to the whole number would be 14.

3 0

2 years ago

Read 2 more answers

A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

2 years ago

What is electrostatic force?

zalisa [80]

"Electrostatic forces are attractive or repulsive forces between particles that are caused by their electric charges."

5 0

2 years ago

You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength

GREYUIT [131]

Answer:

Period of the signal.

Explanation:

So, this question is all about a concept in physics or astronomy which is called or known as Radiation Astronomy and Galactic Nuclei that are active. This concept talks most about Quasars; a powerful radiating object which derives its power from black holes.

When You take a look at Quasars, we get the to know that the more you think you can see, the more they move away from us.

Thus, when "You are observing the radiation from a distant active galaxy and you notice that the amplitude of the signal varies in strength regularly over a certain period. The maximum possible size for the source of this radiation can now be calculated from the "PERIOD OF THE SIGNAL.

NB: not the amplitude but the period.

7 0

3 years ago