Is this right? Please help me ITS SOCIOLOGY

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .

pickupchik [31]

The thermal energy that is generated due to friction is 344J.

<h3>What is the thermal energy?</h3>

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 * 2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

Learn more about thermal energy due to friction:brainly.com/question/7207509

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7 0

2 years ago

What physical quantity is a measure of the amount of inertia an object has?

satela [25.4K]

Mass is the physical quantity

3 0

4 years ago

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

5 0

3 years ago

in the same viewing window. Compare the magnitudes of f (x) and g(x) when x is close to 0. Use the comparison to write a short p

Ivanshal [37]

Answer:

Due to inertia of restttttttrestrestrestrest

6 0

3 years ago

A drop

exis [7]

Answer:

$27.5\ m$

Explanation:

As we know that volume of cylinder is

$v=\pi r^{2} *h$

Where v=volume , h= height or thickness and r= radius

Here,

$v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5$

Putting these values in the previous equation , we get

$10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m$

Therefore thickness is 27.5 m

7 0

3 years ago