PLEASE HELP!! Recreation that uses the large muscles of the body is sometimes referred to as ____

What factors determine the presence of electrical energy?

luda_lava [24]

One of them would be power source.

8 0

4 years ago

A solid sphere of radius 40.0 cm has a total positive charge of 16.2 μC uniformly distributed throughout its volume. Calculate t

Jobisdone [24]

Answer:

(a) E=0 : 0 cm from the center of the sphere

(b) E= 227.8*10³ N/C : 10.0 cm from the center of the sphere

(c)E= 911.25*10³ N/C : 40.0 cm from the center of the sphere

(d)E= 411.84 * 10³ N/C : 59.5 cm from the center of the sphere

Explanation:

If we have a uniform charge sphere we can use the following formulas to calculate the Electric field due to the charge of the sphere

$E=\frac{K*Q}{r^{2} }$ : Formula (1) To calculate the electric field in the region outside the sphere r ≥ a

$E=k*\frac{Q}{a^{3} } *r$ :Formula (2) To calculate the electric field in the inner region of the sphere. r ≤ a

Where:

K: coulomb constant

a: sphere radius

Q: Total sphere charge

r : Distance from the center of the sphere to the region where the electric field is calculated

Equivalences

1μC=10⁻⁶C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

Q=16.2 μC=16.2 *10⁻⁶C

a= 40 cm = 40*10⁻²m = 0.4m

Problem development

(a)Magnitude of the electric field at 0 cm :

We replace r=0 in the formula (2) , then, E=0

(b) Magnitude of the electric field at 10.0 cm from the center of the sphere

r<a , We apply the Formula (2):

$E=9*10^{9} *\frac{16.2*10^{-6} }{0.4^{3} } *0.1$

E= 227.8*10³ N/C

(c) Magnitude of the electric field at 40.0 cm from the center of the sphere

r=a, We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.4^{2} }$

E= 911.25*10³ N/C

(d) Magnitude of the electric field at 59.5 cm from the center of the sphere

r>a , We apply the Formula (1) :

$E=\frac{9*10^{9}*16.2*10^{-6} }{0.595^{2} }$

E= 411.84 * 10³ N/C

4 0

3 years ago

3 نقطة (نقاط)

Molodets [167]

The work-energy theorem says that the total work done on the block is equal to the difference of its kinetic energies at points B and A. Then the total work done on the block is

$W_{\rm total} = K_B - K_A = 4.0\,\mathrm J - 5.0\,\mathrm J = -1.0\,\mathrm J$

Friction acts on the block to oppose its motion, so it does negative work on the block, -4.5 J.

The only other force acting on the block as it moves is the force P. Let $W_P$ be the work done by the force P. Then the total work done on the block is

$W_{\rm total} = W_P + W_{\rm friction} \iff -1.0\,\mathrm J = W_P - 4.5 \,\mathrm J \implies W_P = \boxed{3.5\,\mathrm J}$

5 0

3 years ago

The total weight of a car with passengers is 30,000N. The area of contact of each of the four tyres with the ground is 0.025m2.

vlabodo [156]

Answer:

The minimum pressure car tire pressure is 300,000 Pa

Explanation:

Pressure

It's a measure of the force exerted by an object per unit surface:

$\displaystyle P=\frac{F}{A}$

Where F is the force applied on a surface area A.

We are given the total weight of a car with passengers as Ft=30,000 N and we also know the area of contact of each tire with the ground as $A=0.025 m^2$ .