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Svetllana [295]

2 years ago

13

Which graph shows acceleration? A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds 0

meters upward. A graph of position (meters) versus time (seconds) has a concave line running from 0 seconds 0 meters upward. A graph of position (meters) versus time (seconds) has a straight line running from 0 seconds positive number of meters downward to some later time 0 meters.

2 answers:

anzhelika [568]2 years ago

7 0

Answer:

its the middle one

Explanation:

i did it on edge 2021

charle [14.2K]2 years ago

6 0

It's the one that has a concave line running upward.

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Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

2 years ago

A 0.12 kg bird is flying at a constant speed of 7.8 m/s. what is the birds conetic energy?

lana [24]

KE=1/2mv^2 - equation for kinetic energy
KE=(1/2)(0.12 kg)((7.8 m/s)^2 - plug it into the formula
KE=(0.06 kg)(60.84 m/s) - multiply 1/2 to the mass and square the speed
KE= 3.7 J - answer
Hope this helps

7 0

3 years ago

Read 2 more answers

liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i

aksik [14]

$4.80 \times 10^3 \text { seconds }$ long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

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3 0

1 year ago

Which of the following is a scalar quantity?

neonofarm [45]

Answer:

55

Explanation:

I hope this answer help u

4 0

2 years ago

Read 2 more answers

A transformer has two sets of coils, the primary with N1 = 160 turns and the secondary with N2 = 1400 turns. The input rms volta

vovikov84 [41]

To solve the problem it is necessary to apply the concepts related to the voltage in a coil, through the percentage relationship that exists between the voltage and the number of turns it has.

So things our data are given by

$N_1 = 160$

$N_2 = 1400$

$\Delta V_{1rms} = 62V$

PART A) Since it is a system in equilibrium the relationship between the two transformers would be given by

$\frac{N_1}{N_2} = \frac{\Delta V_{1rms}}{\Delta V_{2rms}}$

So the voltage for transformer 2 would be given by,

$\Delta V_{2rms} = \frac{N_2}{N_1} \Delta V_{1rms}$

PART B) To express the number value we proceed to replace with the previously given values, that is to say

$\Delta V_{2rms} = \frac{N_1}{N_2} \Delta V_{1rms}$

$\Delta V_{2rms} = \frac{1400}{160} 62V$

$\Delta V_{2rms} = 1446.66V$

7 0

3 years ago