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inysia [295]
3 years ago
13

An acorn with a mass of .0300 kg is hanging from a branch in a tree it is 2.50 M off the ground what is the potential energy of

the acorn
0.735 j
0.075 j
1.84 j
0.00j
Physics
2 answers:
Alekssandra [29.7K]3 years ago
5 0
According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m ) 
</span>Hope it helps
Gennadij [26K]3 years ago
5 0
0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m ) 
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Explain why the extrapolated temperature is used to determine the maximun temperature of the mixture rather than the highest rec
Oksi-84 [34.3K]
The extrapolated temperature is used to define the maximum temperature of the mixture relatively than the highest recorded temperature in which the conclusion will effect in a higher specific heat value. Heat is bound to escape from whatever apparatus is using, therefore it is needed to account for the loss of the heat that does not go into increasing the temperature of the mixture. 
6 0
3 years ago
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What is the sum of 231 + 5080 + 80 + 13 + 9​
Gemiola [76]

Answer:

5413

Explanation:

add them up

6 0
3 years ago
A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface
Fiesta28 [93]

Answer:

X_2=25.27m

Explanation:

Here we will call:

1. E_1: The energy when the first spring is compress

2. E_2: The energy after the mass is liberated by the spring

3. E_3: The energy before the second string catch the mass

4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

2. E_2 -E_3= W_f

3.E_3 = E_4

where W_f is the work of the friction.

1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

where K is the constant of the spring, x is the distance compressed, M is the mass and V_2 the velocity, so:

\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

solving for X_2, we get:

X_2=25.27m

3 0
3 years ago
A small 20-kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?
MariettaO [177]
Ek=1/2mv^2 = (20)(1/2)(2x2) =40
8 0
3 years ago
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The human tibia breaks under an impulse of 55 Ns. If after falling a person typically comes to eat in a time span of 0.005 s how
ozzi

Answer:

0.73 m/s

Explanation:

From Newton second law of motion,

I = m(v-u)...................... Equation 1

Where I = Impulse, m = mass of the person, v = final velocity, u = Initial velocity.

make v the subject of the equation

v =(I/m)+u................. Equation 2

Note: u = 0 m/s as the person is falling from an height.

Given: I = 55 Ns, m = 75 kg, u = 0 m/s

Substitute into equation 2

v = 55/75

v = 0.73 m/s

5 0
4 years ago
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