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3 years ago

Newton's first law of motion is the law of inertia. What does it state?

Physics

1 answer:

vladimir2022 [97]3 years ago

5 0

Answer:

b is the right answer

Explanation:

<h3>hope helps!!</h3>

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g A high-energy photon turns into and electron and a positron. (A positron has exactly the same mass as the electron, but opposi

yawa3891 [41]

Answer:

2 m = E / c^2 where m is mass of electron

E = h v where v is the frequency ( nu) of the incident photon

E = h c / y where y is the incident wavelength (lambda)

2 m = h / (c y)

y = h / (2 m c) wavelength required

y = 6.62 * 10E-34 / (2 * 9.1 * 10E-31 * 3 * 10E8) m

y = 3.31 / 27.3 E-11 m

y = 1.21 E -12 m = .0121 Angstrom units

5 0

3 years ago

Just about everyone at one time or another has been burned by hot water or steam. This problem compares the heat input to your s

tatyana61 [14]

Answer:

$Q_T=63313.5\ J$

Explanation:

Given:

temperature of skin, $T_s=34^{\circ}C$

initial temperature of steam vapour, $T_v=100^{\circ}C$

latent heat of steam, $L=2256\ J.g^{-1}$

mass of steam, $m=25\ g$

specific heat of water, $c=4190\ J.kg^{-1}.K^{-1}=4.19\ J.g^{-1}.K^{-1}$

final temperature, $T_f=34^{\circ}C$

<em>Assuming that no heat is lost in the surrounding.</em>

$Q=m.c.\Delta T$

<u>Now the total heat given by the steam to form water at the given conditions:</u>

$Q_T=Q_{Lv}+Q_w$ ..............................(1)

where:

$Q_{Lv}=$ latent heat given out by vapour to form water of 100°C

$Q_w=$ heat given by water of 100°C to come at 34°C.

putting respective values in eq. (1)

$Q_T=m(L+c.\Delta T)$

$Q_T=25(2256+4.19\times 66)$

$Q_T=63313.5\ J$

is the heat transferred to the skin.

4 0

3 years ago

In the areas called pauses such as the tropopause, the temperature line is vertical. What does this indicate about the temperatu

ivolga24 [154]

Answer:

the answer is A

Explanation: The vertical temperature line indicates no change in temperature at those altitudes.

6 0

3 years ago

Read 2 more answers

In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves wit

Minchanka [31]

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

$x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}$

Putting, t = 1 s and x = 4 m in above equation, we get :

$4 = \dfrac{a(1)^2}{2}\\\\a = 8 \ m/s^2$

Therefore, the acceleration of the cart is 8 m/s².

5 0

3 years ago

Which form of heat transfer occurs when objects of two temperatures touch

Arada [10]

Conducting because it occurs when two objects touch and heat is transferred

7 0

3 years ago