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ipn [44]
3 years ago
13

Boost your own ______

Physics
1 answer:
Leni [432]3 years ago
6 0

Answer:

Self-confidence

Explanation:

Boost your own self-confidence by living your life, not watching others live their fake ones!

The other choices do not fit as well as self-confidence does, but it's also a common phrase to say "boost your own self-confidence."

You might be interested in
A block of ice (m = 9 kg) at a temperature of T1 = 0 degrees C is placed out in the sun until it melts, and the temperature of t
jonny [76]

Answer:

a) An expression for the amount of energy, E_m, needed to melt the ice into water.

(E_m) = (m × Lf)

b) An expression for the total amount of energy, E_tot, to melt the ice and then bring the water to T2

(Total heat) = (m × Lf) + mc (T2 - T1)

c) 3,646,458 J = 3646.46 kJ

Explanation:

a) When a pure body changes its phase at meltimgbor boiling point, it does so at a constant temperature. When a pure body melts, the amount of heat responsible for this change is just given by a product od the mass of the body and the body's heat of fusion.

(E_m) = (m × Lf)

b) The Heat required to raise the temperature of a body from one temperature to another is given by the product of the mass of the body, its specific heat capacity and the temperature difference between the final point and the starting point.

(E_2) = mcΔT = mc (T2 - T1)

Total heat required to melt the ice at T1 = 0 and raise the temperature of the resulting water to T2 is then a sum of (E_m) + (E_2)

(Total heat) = (m × Lf) + mc (T2 - T1)

c) What is the energy in Joules?

(Total heat) = (m × Lf) + mc (T2 - T1)

m = mass of ice = resulting mass of water = 9 kg

Lf = latent heat of fusion = 334000 J/kg

c = Specific heat capacity of water = 4186 J/kg.K

T2 = final temperature of the water = 17°C

T1 = Initial temperature of the water = 0°C

Note that the units of temperature difference is the same for K and °C

(Total heat) = (m × Lf) + mc (T2 - T1)

Q = (9 × 334000) + [9 × 4186 × (17 - 0)]

Q = 3,006,000 + 640,458 = 3,646,458 J = 3646.46 kJ

Hope this Helps!!!

7 0
4 years ago
Can you please help me find the answer
Serggg [28]
Im sorry may you please retake the picture then i will answer
8 0
4 years ago
The ratio of output power to input power, in percent, is called?
Harrizon [31]
That's efficiency. There's no law that it must be stated in percent.
4 0
3 years ago
Read 2 more answers
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
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