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3 years ago

How were the fossil symbols and mountain belts helpful in deciding where to move the continents

1 answer:

6 0

Hey there!

The fossil symbols and mountain belts provided clues as to where and how the continents were connected together before they had moved. The mountain belts showed scientists the direction those plates moved and the fossils indicated that similar dinosaurs or animals lived together in the same area.

Thank you!

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A basketball rolls to a stop along a gym floor without anyone touching it. What answer choice BEST explains why this occurred?

Minchanka [31]

Frictional forces are exerting force against the basketball, causing it to stop.

7 0

3 years ago

Read 2 more answers

At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp

motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

$K = \frac{1}{2} mv^2$

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51 = 1.07 kg

The mass of the body at that instant is 1.07 kg

5 0

3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

3 years ago

1) Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way ba

timurjin [86]

Answer:

<h3>B. 19miles</h3>

Explanation:

If Freddy drives 4 miles east to his friend's house. He then travels 9 more miles east to the supermarket. Finally on his way back home he out of gas 6 miles after leaving the supermarket, the distance travel by fred will be the sup of all the distances he covered throughout the journey.

Distance covered by fred = 4miles + 9miles + 6miles

Distance covered by fred = 13miles + 6miles

Distance covered by fred = 19miles

8 0

3 years ago

A projectile is fired with an initial velocity of 120.0 meters per second at an angle, θ above the horizontal. If the projectile

k0ka [10]

Answer:

θ = 62.72°

Explanation:

The projectile describes a parabolic path:

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x = x₀+ vx*t Formula (1)

vx = v₀x

Where:

x: horizontal position in meters (m)

x₀: initial horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m/s

v₀x: Initial speed in x in m/s

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis are:

y= y₀+(v₀y)*t - (1/2)*g*t² Formula (2)

vfy= v₀y -gt Formula (3)

Where:

y: vertical position in meters (m)

y₀ : initial vertical position in meters (m)

t : time in seconds (s)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Data

v₀ = 120 m/s , at an angle θ above the horizontal

v₀x= 55 m/s

x-y components of the initial velocity ( v₀)

v₀x = v₀*cosθ Equation (1)

v₀y = v₀*sinθ Equation (2)

Calculating of the angle θ

We replace data in the Equation (1)

55 = 120*cosθ

cosθ = 55 / 120

$\theta = cos^{-1}( \frac{55}{120} )$

θ = 62.72°

3 0

2 years ago