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IrinaK [193]
3 years ago
15

Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slid

e. If they both start at the same height above the ground, which of the following is true about their kinetic energies as they reach the ground?
A) Bob's kinetic energy is greater than Esther's.B) Esther's kinetic energy is greater than Bob's.C) They have the same kinetic energy.D) The answer depends on the shape of the slide.
Physics
1 answer:
galben [10]3 years ago
5 0

Answer:

Explanation:

They have the same kinetic energy

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Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?
Alina [70]
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
F= k_e \frac{ q_1 q_2}{r^2}
where k_e=8.99\cdot10^9~N m^2 C^{-2} is the Coulomb's constant, q_1=q_2=29~nC=29\cdot 10^{-9}~C is the intensity of the two charges, and r=5.5~cm=0.055~m is the distance between them.

Substituting these numbers into the equation, we get
F=2.5~10^{-3}~N

The force is repulsive, because the charges have same sign and so they repel each other.
6 0
3 years ago
The laboratory test that detects neutropenia is
bixtya [17]
It's absolute neutropihil count, also called ANC
4 0
4 years ago
When your sled starts down from the top of a hill, it hits a frictionless ice slick that extends all the way down the hill. At t
Blababa [14]

Answer:

Explanation:

Given that,

Height of hill is 50m

Coefficient of friction is μ=0.62

Energy is conserved, then

K.E at the bottom of the hill is equally to P.E at the top of the hill

½mv²=mgh

Mass cancel out

½v²=gh

v²=2gh

v=√2gh

Since g=9.81 and h=50

v=√2×9.81 ×50

v=31.32m/s

This is the initial speed at the bottom of the hill

At the bottom of the hill 3 forces are acting on the body

1. Weight,

2. Normal

3. Frictional force

Now taking newton law of motion

ΣF = ma.

Along y axis, since the body is not moving in y direction, they acceleration along y is 0m/s²

ΣFy = 0

N-W=0

N=W

Since weight =mg

N=W=mg

Using law of friction, Fr=μN

Therefore,

Fr=μmg

Applying Newton law to the x-direction

ΣFx = ma

Fr=ma,

Since Fr=μ mg

μ mg=ma

a=μg

Given that μ=0.62 and g=9.81

a=0.62×9.82

a=6.076m/s²

Since we have acceleration, we can use any of the equation of motion to find distance travel

v²=u²-2gS, . this is due that the box is decelerating and it comes to a halt and this show that final velocity v=0m/s

Then,

0²=31.32²-2×9.81 ×S

-31.32² =-2×9.81×S

S=31.32²/(2×9.81)

S=50.05m

The sled will travel a distance of 50.1m once it reach the bottom

5 0
3 years ago
Residential building codes typically require the use of 12-gauge copper wire (diameter 0.2053 cm) for wiring receptacles. Such c
AysviL [449]

Answer:

a) E = 4.26 W

b) E' = 6.724 W

c) copper wire is the safer option to use.

Explanation:

Given:

Diameter of the 12 gauge copper wire, d = 0.2053 cm

Thus, Radius of the 12 gauge copper wire, r = 0.2053 cm

/ 2 = 0.10265 cm  = 0.10265 × 10⁻² m

Now.

the area (A) comes out as

A = π × (0.10265 × 10⁻²)²

A = 3.3103 × 10⁻⁶ m²

Length of the copper wire, L = 2.10 m

a) The resisitivity (ρ) of copper = 1.68 × 10⁻⁸ ohm m

Now,

the resistance of the copper , R = ρL/A

or

R = (1.68 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.01065 ohm

The Energy (E) is given as,

E = I²R

where, I is the current

I = 20.0 A

on substituting the values, we get

E = 20.0² × 0.01065

E = 4.26 W

(b) For the aluminium

Resisitivity, ρ' = 2.65 × 10⁻⁸  ohm m

Now, the resistance of the aluminium wire, R' = (ρ' × L) / A

Since the cross-section of the aluminium wire is same as the copper wire

thus,

R = (2.65 × 10⁻⁸ × 2.1) / ( 3.3103 × 10⁻⁶)

or

R = 0.0168 ohm

Therefore,

The Rate of energy produced by the aluminium wire, E' = I²R'

or

E' = 20.0² ×  0.0168

or

E' = 6.724 W

(c) From the above results, we can conclude that the power consumed or the rate of energy produced by the aluminium wire is more.

Hence, copper wire is the safer option to use.

8 0
4 years ago
Which describes the relation between an electric field and an equipotential surface?
alina1380 [7]

The answer is the field vector is perpendicular to the surface.

The electric field's angle with the equipotential surface is always 90 degree. The electric field is always perpendicular to the equipotential surface.

What is electric field?

  • The electric field is formally defined as a vector field associated with each location in space, the force per unit charge exerted on a positive test charge at rest at that place.
  • The electric charge or time-varying magnetic fields create the electric field. At the atomic level, the electric field is responsible for the attractive forces that hold the atomic nucleus and electrons together.
  • The normal vector to a surface, often known as the "normal," is a vector that is perpendicular to the surface at a particular position. When considering normals on closed surfaces, the inward-pointing normal (pointing towards the surface's interior) and outward-pointing normal are commonly differentiated.

To learn more about electric field visit:

brainly.com/question/8971780

#SPJ4

8 0
2 years ago
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