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Lyrx [107]
3 years ago
15

Obtain the temp. at which the reading of Fahrenheit is double of that in Celsius scale.​

Physics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer:

320 degrees fahrenheit and 160 degrees celcius

Explanation:

there's a formula that i definately don't remember but it exists

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Answer:

Yes, it is correct : )

Explanation:

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Describe the difference between the currents that exist in the wires leading to a capacitor when these wires are connected to (a
krek1111 [17]

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Explanation: In DC circuit, the current will flow for a short time, which is required to charge the capacitor. Once you switch it on, it spikes and the gradually decreases to almost zero (0) as the capacitor becomes fully charged.

In an AC circuit, the circuit acts as if the current is flowing throw the plates whereas is not actually flowing. The circuit acts like the AC is flowing through the capacitor.

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1 hp is equal to? What is the answer​
rewona [7]
The electrical equivalent of one horsepower is 746 watts in the International System of Units (SI), and the heat equivalent is 2,545 BTU (British Thermal Units) per hour. Another unit of power is the metric horsepower, which equals 4,500 kilogram-metres per minute (32,549 foot-pounds per minute), or 0.9863 horsepower.
5 0
3 years ago
For general projectile motion, the horizontal component of a projectile's velocity aSelect one:O a. continuously increases.O b.
liubo4ka [24]

ANSWER:

d. remains a non-zero constant.

STEP-BY-STEP EXPLANATION:

If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant

3 0
1 year ago
A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
3 years ago
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