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yarga [219]
3 years ago
13

Alexandra drops an egg from 30 m above the ground to hit Vanessa on the head. If Vanessa stands 3.5 meters tall. How fast is the

egg moving when it strike her head? (Hint: Draw a picture and you will see why you have to consider her height and find the time first!)
Physics
1 answer:
I am Lyosha [343]3 years ago
7 0

This is problem of free falling objects, which can be solved using the formula:

V = sqrt(2gy)

Where v is the velocity upon impact

G is the acceleration due to gravity ( 9.81 m/s2)

Y is the height

Since Venessa is 3.5 m

Y = 30 -3.5 = 26.5 m

V = sqrt(2 (9.81 m/s2) ( 26.5 m))

<span>V = 22.8 m/s</span>

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What is energy and types of energy<br>​
pishuonlain [190]
Energy- the ability to do work/how things can change and move

Types
Potential Energy
Kinetic Energy
Nuclear Energy
Mechanical Energy
Sound Energy
Heat
8 0
3 years ago
Read 2 more answers
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
3 years ago
As temperature increases, ________. Group of answer choices the resistance of a conductor remains the same the resistance of a c
amm1812

Answer:

resistance of a conductor increases

Explanation:

The resistance of conductors is directly proportional to the temperature of the conductor. This implies that when the temperature of the conductor is increased, the resistance of the conductor increases likewise.

This is applied in the resistance thermometer. Resistance thermometers are useful for accurate temperature measurements at very high or very low temperatures.

6 0
3 years ago
An airliner arrives at the terminal, and its engines are shut off. The rotor of one of the engines has an initial clockwise angu
Ilia_Sergeevich [38]

(a) 1200 rad/s

The angular acceleration of the rotor is given by:

\alpha = \frac{\omega_f - \omega_i}{t}

where we have

\alpha = -80.0 rad/s^2 is the angular acceleration (negative since the rotor is slowing down)

\omega_f is the final angular speed

\omega_i = 2000 rad/s is the initial angular speed

t = 10.0 s is the time interval

Solving for \omega_f, we find the final angular speed after 10.0 s:

\omega_f = \omega_i + \alpha t = 2000 rad/s + (-80.0 rad/s^2)(10.0 s)=1200 rad/s

(b) 25 s

We can calculate the time needed for the rotor to come to rest, by using again the same formula:

\alpha = \frac{\omega_f - \omega_i}{t}

If we re-arrange it for t, we get:

t = \frac{\omega_f - \omega_i}{\alpha}

where here we have

\omega_i = 2000 rad/s is the initial angular speed

\omega_f=0 is the final angular speed

\alpha = -80.0 rad/s^2 is the angular acceleration

Solving the equation,

t=\frac{0-2000 rad/s}{-80.0 rad/s^2}=25 s

6 0
3 years ago
Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th
Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0
3 years ago
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