Calculate the speed of a bus that travels a distance of 55 miles in 0.75 hours.

Scientists often use models to study the movement of continents. Why might scientists use a model to show this movement? A. Extr

egoroff_w [7]

Answer:

Paleontologists have argued for a long time that the demise of the dinosaurs was caused by climatic alterations associated with slow changes in the positions of continents and seas resulting from plate tectonics. Off and on throughout the Cretaceous (the last period of the Mesozoic era, during which dinosaurs flourished), large shallow seas covered extensive areas of the continents. Data from diverse sources, including geochemical evidence preserved in seafloor sediments, indicate that the Late Cretaceous climate was milder than today's. The days were not too hot, nor the nights too cold. The summers were not too warm, nor the winters too frigid. The shallow seas on the continents probably buffered the temperature of the nearby air, keeping it relatively constant.

3 0

2 years ago

Read 2 more answers

Calculate (a) the torque, (b) the energy, and (c) the average power required to accelerate Earth in 4.0 days from rest to its pr

natima [27]

<h2>Answer:</h2>

Torque = <em>2.05 x 10²⁸ Nm</em>

Energy = <em>3.54 x 10³³ J</em>

Average power = <em>1.02 x 10²⁸ W</em>

<h2>Explanation:</h2>

(a) Torque (τ) is the rotational effect of a given force.

It is given by

τ = I x α -------------(i)

Where;

I = rotational inertia of the object

α = angular acceleration of the object.

In this case, the object is the Earth. Therefore,

I = 9.71 x 10³⁷ kg m²

α = ω / t

Where;

ω = angular velocity of earth = 2π rad / day

<em>Since </em>

<em>1 day = 24 hours and 1 hour = 3600seconds</em>

<em>1 day = 24 x 3600 seconds = 86400seconds</em>

<em>=> ω = 2π rad / 86400seconds</em>

<em />

t = 4 days = 4 x 24 x 3600 seconds = 345600 seconds

=> α = ω / t

=> α = 7.29 × 10⁻⁵ / 345600

=> α = (7.29 × 10⁻⁵) / (3.456 x 10⁵)

=> α = (7.29 × 10⁻⁵⁻⁵) / (3.456)

=> α = (7.29 × 10⁻¹⁰) / (3.456)

=> α = 2.11 × 10⁻¹⁰ rad/s²

Now substitute the values of I and α into equation (i)

τ = 9.71 x 10³⁷ x 2.11 × 10⁻¹⁰

τ = 9.71 x 10²⁷ x 2.11

τ = 20.5 x 10²⁷ Nm

τ = 2.05 x 10²⁸ Nm

(ii) The energy (rotational energy) E is given by;

E = $\frac{1}{2}$ x I x ω

E = $\frac{1}{2}$ x 9.71 x 10³⁷ x 7.29 × 10⁻⁵

E = 35.4 x 10³² J

E = 3.54 x 10³³ J

(iii) The average power P, is given by;

P = E / t

P = 3.54 x 10³³ / 345600

P = 1.02 x 10²⁸ W

5 0

3 years ago

PLEASE HELPPPP ME WITH THIS

lozanna [386]

I am not so sure about this it is too difficult

4 0

3 years ago

What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

4 0

2 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago