What is the unit measure for energy

A light ray strikes a flat, smooth, reflecting surface at an angle of 80° to the normal. What angle does

charle [14.2K]

Answer:

80 angle of incidence=angle of reflection

Explanation:

7 0

3 years ago

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An empty beaker is placed on a top-pan balance. Some water is now poured into the beaker.What is the weight of the water? A. 0.0

mario62 [17]

Answer:

A. 0.044 kg

Explanation:

We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.

Hope it is enough.

Please mark me as brainliest.

6 0

3 years ago

A student rides her bike to school. Her school is 5 miles from home. She travels at an average rate of 15 miles per hour. How mu

Elden [556K]

Answer:

.3 repeating hours

Explanation:

3 0

2 years ago

Starting from rest, a 68.0 kg woman jumps down to the floor from a height of 0.790 m, and immediately jumps back up into the air

Aleonysh [2.5K]

Answer:

a) I = 0 N s, b) v = -3.935 m / s, c) vf = 3.935 m / s, d) y = 0.790 m

Explanation:

a) Let's start by defining the upward direction (+ y) as positive. For this part of the exercise we must use the momentum relationship

I = ∫F dt

The force of the woman on the floor is given and by action and rection the floor exerts on the woman a force of equal magnitude, but opposite direction

I = ∫ (9200 t - 11500 t2) dt

I = 9200 t² / 2 - 11500 t³ / 3

We evaluate between the lower limit t = 0 and upper limit t = 0.800 s

I = 9200 (0.8² -0) - 11500 (0.8³ -0)

I = 5888 -5888

I = 0 N s

Directed from the floor to the woman

b) For this part we use kinematics

v² = v₀² - 2g y

v = √ (0 - 2 9.8 (-0.79))

v = 3.935 m / s

The speed direction is down

c) for this we use the relationship between momentum and the amount of movement

I = ΔP

I = m vf - m v₀

vf = (I + m v₀) / m

This is the impulse of women on the floor

vf = ( 0 + 68 (3.935)) / 68

vf = 3.935 m / s

d) let's use kinematics

v₂ = v₀² - 2gy

0 = v₀² - 2gy

y = v₀² / 2g

y = 3.935²/2 9.8

y = 0.790 m

8 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced: