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iogann1982 [59]
3 years ago
7

A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ri

ng to pull the trampoline slowly down while his 79-kg mother is sleeping on it. When he releases the trampoline, she is launched upward. As she passes through the position at which she was before her son stretched the trampoline, her speed is 2.5 m/s .
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
5 0

Answer:

Explanation:

Given

Mass of mother m=79\ kg

Mother is released with a speed of v=2.5\ m/s

Assuming we need to find elastic Potential energy stored in Trampoline

Kinetic energy acquired  by Mother comes from Elastic Potential Energy stored in the spring

I.e. Elastic Potential Energy=Kinetic Energy of mother

\frac{1}{2}kx^2=\frac{1}{2}mv^2

E=\frac{1}{2}\times 79\times 2.5^2=246.875\ J

So 246.875 J of Energy is stored in the trampoline as he pull the ring

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1. How much energy is needed to raise 1 g of water 1°C?
Serga [27]

Answer:

1.The calorie was originally defined as the amount of heat required at a pressure of 1 standard atmosphere to raise the temperature of 1 gram of water 1° Celsius. Since 1925 this calorie has been defined in terms of the joule, the definition since 1948 being that one calorie is equal to approximately 4.2 joules.

2.Boiling water at 100 degrees Celsius: 540 calories are needed to turn 1 gram (at 100 degrees Celsius) of water to steam.

5 0
3 years ago
the ocean floor is, on average, 4267 m below sea level. What is the pressure in the atmosphere at this depth?
hichkok12 [17]
The pressure at the depth h in the ocean is given by (Stevin's law)
p= p_0 + \rho g h
where
p_0 = 1.0 \cdot 10^5 Pa is the atmospheric pressure
and \rho g h is the pressure exerted by the column of water of height h=4267 m, with \rho = 1000 kg/m^3 being the water density and g=9.81 m/s^2.
Substituting, we find
p=1.0 \cdot 10^5 Pa + (1000 kg/m^3)(9.81 m/s^2)(4267 m)=4.20 \cdot 10^7 Pa
We want to convert this into atmospheres: we know that 1 atm corresponds to the atmospheric pressure at sea level, so 1 atm=1.0\cdot 10^5 Pa, therefore we just need to divide by this number:
p= \frac{4.20 \cdot 10^7 Pa}{1.0 \cdot 10^5 Pa/atm} =420 atm
7 0
2 years ago
Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie
elena-14-01-66 [18.8K]
The electric field generated by a point charge is given by:
E= k_e \frac{Q}{r^2}
where
k_e = 8.99 \cdot 10^9 Nm^2 C^{-2} is the Coulomb's constant
Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them. 

Let's calculate first the electric field generated by the positive charge at that point:
E_1=k_e  \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C
where the negative sign means its direction is toward the charge.

If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
E_{tot} =E_1 + E_2 = 7.19 \cdot 10^6 N/C+7.19 \cdot 10^6 N/C=1.44 \cdot 10^7 N/C
3 0
3 years ago
A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen
Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

a=\dfrac{v^2}{R}

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

a'=\dfrac{v^2}{R'}

a'=\dfrac{v^2}{R/4}

a'=4\times \dfrac{v^2}{R}

a'=4\times a

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0
3 years ago
Two objects must be in contact for them to exert a force on each other. <br> True or False
nevsk [136]

Answer:

False

Explanation:

This proposition is false because by example the sun exerts a force over the earth and them are not in contact

7 0
2 years ago
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