All categories

2 years ago

Select which geologic events will have an immediate impact on the landscape. Choose three (3).

Physics

1 answer:

alexgriva [62]2 years ago

6 0

Answer: A, B, and D,

C is incorrect because continental drift takes several centuries to move the tectonic plates witch is not an immediate impact leaving you with A,B, and D.

You might be interested in

An object is thrown into the air at 60m/s, straight up. What is its velocity at the highest point?

Cloud [144]

Explanation:

Whenever an object is at its highest point, the velocity and acceleration of the object is zero.

6 0

3 years ago

Two bodies fall freely from different heights and reach the ground simultaneously. The time of descent for the first body is 1s

jok3333 [9.3K]

The initial height of the first body is given by:
$h_1 = \frac{1}{2}gt^2$
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
$h_1 = \frac{1}{2}(9.81 m/s^2)(1 s)^2=4.9 m$

The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
$h_2 = \frac{1}{2}(9.81 m/s^2)(2 s)^2=19.6 m$

The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.

6 0

3 years ago

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stoc

vivado [14]

Answer:

Δ $R_{e}$ = 84 Ω, $R_{e}$ = (40 ± 8) 10¹ Ω

Explanation:

The formula for parallel equivalent resistance is

1 / $R_{e}$ = ∑ 1 / Ri

In our case we use a resistance of each

R₁ = 500 ± 50 Ω

R₂ = 2000 ± 5%

This percentage equals

0.05 = ΔR₂ / R₂

ΔR₂ = 0.05 R₂

ΔR₂ = 0.05 2000 = 100 Ω

We write the resistance

R₂ = 2000 ± 100 Ω

We apply the initial formula

1 / $R_{e}$ = 1 / R₁ + 1 / R₂

1 / $R_{e}$ = 1/500 + 1/2000 = 0.0025

$R_{e}$ = 400 Ω

Let's look for the error (uncertainly) of Re

$R_{e}$ = R₁R₂ / (R₁ + R₂)

R’= R₁ + R₂

$R_{e}$ = R₁R₂ / R’

Let's look for the uncertainty of this equation

Δ $R_{e}$ / $R_{e}$ = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’

The uncertainty of a sum is

ΔR’= ΔR₁ + ΔR₂

We substitute the values

Δ $R_{e}$ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)

Δ $R_{e}$ / 400 = 0.1 + 0.05 + 0.06

Δ $R_{e}$ = 0.21 400

Δ $R_{e}$ = 84 Ω

Let's write the resistance value with the correct significant figures

$R_{e}$ = (40 ± 8) 10¹ Ω

6 0

3 years ago

Assume that the temperature, 40C, and volume of the gas, 6 liters, remain constant. At 2 atmosphere of pressure, there are 0.036

MAVERICK [17]

Answer:PV=nRT

Ideal gas equation

Explanation:

PV=nRT

Ideal gas equation

7 0

3 years ago

When the magnetic flux through a single loop of wire increases by , an average current of 40 A is induced in the wire. Assuming

Zielflug [23.3K]

COMPLETE QUESTION:

When the magnetic flux through a single loop of wire increases by 30 Tm^2 , an average current of 40 A is induced in the wire. Assuming that the wire has a resistance of 2.5 ohms , (a) over what period of time did the flux increase? (b) If the current had been only 20 A, how long would the flux increase have taken?

Answer:

(a). The time period is 0.3s.

(b). The time period is 0.6s.

Explanation:

Faraday's law says that for one loop of wire the emf $\varepsilon$ is

$(1). \: \: \varepsilon = \dfrac{\Delta \Phi_B}{\Delta t }$

and since from Ohm's law

$\varepsilon = IR$ ,

then equation (1) becomes

$(2). \: \:IR= \dfrac{\Delta \Phi_B}{\Delta t }.$

(a).

We are told that the change in magnetic flux is $\Phi_B = 30Tm^2$ , the current induced is $I = 40A$ , and the resistance of the wire is $R = 2.5\Omega$ ; therefore, equation (2) gives

$(40A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

which we solve for $\Delta t$ to get:

$\Delta t = \dfrac{30Tm^2}{(40A)(2.5\Omega)},$

$\boxed{\Delta t = 0.3s}$ ,

which is the period of time over which the magnetic flux increased.

(b).

Now, if the current had been $I =20A$ , then equation (2) would give

$(20A)(2.5\Omega)= \dfrac{30Tm^2}{\Delta t },$

$\Delta t = \dfrac{30Tm^2}{(20A)(2.5\Omega)},$

$\boxed{\Delta t = 0.6 s\\}$

which is a longer time interval than what we got in part a, which is understandable because in part a the rate of change of flux $\dfrac{\Delta \Phi_B}{\Delta t}$ is greater than in part b, and therefore , the current in (a) is greater than in (b).

7 0

3 years ago